Welcome back to the lecture on Quantum Chemistry or Principles of Quantum Mechanics. This is the third lecture giving some more details on the mathematics basic mathematics particularly matrices and the algebra of matrices. In the last lecture we looked at the Eigen values and Eigen vectors of a simple 2 by 2 matrix and then I gave you some examples to work out as problems. We shall continue that in this lecture since Eigen values and Eigen vectors are fundamentally important quantities in quantum mechanics. The Schrodinger equation particularly the time independent Schrodinger equationwhich commonly people read as H psi is equal to E psi where H is the Hamiltonian and psi is the wave function and E the energy Eigen value. The Schrodinger equation is an Eigen value equation for the Hamiltonian matrix or the Hamiltonian operator and the size or Eigen functions of the Hamiltonian operator and a major activity in quantum mechanics is to determine the Eigen functions once the Hamiltonian is known. And therefore, the elementary properties of Eigen values and Eigen vectors that we examined we examined in the last lecture and the ones we will look at in this lecture and probably later on these are extremely important for you to keep in mind as they will come again and again ok. So, let me start with today's lecture. So, we are on to the third lecture. So, lecture 3 properties or I will say more properties of matrices I am Mangala Sundar from the Chemistry Department of IIT Madras. This lecture is brought to you through the funding and the support of the National Program on Technology Enhanced Learning organized by the IIT's and Indian Institute of Science and funded by the Ministry of Human Resource Development. So, again like the last two lectures we shall divide this into some two or three smaller parts and in the first part of the lecture which is about say 15 to 20 minutes I shall talk about a couple of elementary properties as well as definitions that we have to worry about. So, let us look at the properties of the traces, determinants and some simple matrices like the orthogonal unitary and Hermitian matrices. We shall follow this with specific properties of Hermitian matrices because Hermitian matrices are the most directly useful quantities in the quantum mechanics particularly the Hamiltonian matrix and any other matrix of operator which is an observable or an experimentally measurable quantity those matrices are all Hermitian. So, we shall study the properties of Hermitian matrices a little more only to the point that we need them and not worry about the formal details. So, first we will do look at the traces and determinants and after doing this I shall give you the example of the eigenvalues and eigenvectors of a 3 by 3 matrix that I promised in the last lecture, but these properties are important and I have sort of put them in between the eigenvalues of a 2 by 2 matrix and eigenvalues of 3 by 3 matrix. So, let us look at the trace, if there are 2 matrices a and b such that a b is not equal to b a that is they do not commute we still have the trace of a b is equal to the trace of b a. It is easy to verify because suppose we write a by the elements a i j and b by the elements b i j then a b matrix element i j would be the sum over the column index k or row index as you see it a i k b k j. The trace of a b is nothing, but the sum of the diagonal elements of a b a b i i and therefore, we can immediately write that the trace of a b is given by sum over k it is the i i matrix element. So, instead of j we have i. So, it is a i k b k j, but summed over all i as well k i k i k k i, but the these being numbers a i k and b k i being complex numbers or real numbers or whatever we can rewrite to this as i k b k i a i k and this is easily seen as sum over k sum over i b k i a i k. And you can write immediately that this particular sum is the matrix element k k of the matrix b a k k and what we have is the sum over k. Therefore, this is the trace of b a. So, this is an extremely important property namely trace of two matrices a b is equal to the trace of a. From this it follows immediately that when we have more than two matrices we should follow a cyclic order namely if you want to write trace a b c three matrices then there are different ways of associating them. So, if you do a and b c as another matrix then this is the same as the trace of this b c matrix which is a product times a by exactly this interchange. The result of b c is a matrix anyway. So, a times that is the same as that times a one relation. Now, instead of a b c if you do the following trace of a b c and write this as trace of a b the result of a b being a matrix times c that is the same as trace of c times a b. And therefore, you see that the trace of a b c is cyclically invariant namely it is equal to trace of b c a in that order a b c order and that is equal to trace of c a p. Now, this is a very simple way of looking up to the trace property of products of matrices, but they are cyclically invariant except when a and b that is the only case that a b is the trace of a b is equal to the trace of b a and if a and b commute the product a b itself is equal to the product b a. So, we do not we have a trivial identity that the trace of the matrix is equal to the trace of itself I mean it does not need anything beyond that, but if the matrices do not commute this property is still important ok. In the same way we can talk about the determinant of the two matrices a b is equal to the determinant of b times a ok. And you can see that this gives rise to the same cyclical invariance for the determinant of the product of many matrices n if you do that it is determinant of b c n times a which is the same as determinant of you repeat this several times you will get n a b c dot. We made use of these two properties in the eigenvalues and eigenvectors determination in the last lecture when we said that the sum of the eigenvalues is the same as the trace of the original matrix and the product of the eigenvalues of the matrix is the same as the determinant of the original matrix. I mean there we did make use of these cyclical relations, but we see where they come from basically from the elementary definition of matrix multiplication. Now, the other part that we need to worry about is two more special matrices called orthogonal and unitary matrices. A matrix A is orthogonal if its inverse is given by the transpose of the matrix. This means that the determinant of A is never 0 because the inverse exists the transpose of any matrix can be obtained by transposing the rows and columns. Therefore, if that has to be equal to the inverse of the matrix obviously, the determinant of that matrix has to be non 0, but if you look at it carefully when you write A A t which is the same as A A inverse matrices ok. If A inverse is equal to A t then A A t is equal to the identity matrix. Therefore, determinant of A A t is equal to the determinant of 1 which is of course, 1 the identity matrix. So, what you have is you have the determinant of A squared please remember A t the determinant of A t is the same as the determinant of A therefore, A A t is the same as the determinant A whole square ok or is equal to the determinant of 1 which is identity which implies that the determinant of A has to be either plus 1 or minus 1 and the case of plus 1 A is known as special orthogonal matrix. Now, if we examine this a little bit further when we write A A t is equal to 1 that is the identity matrix what we imply is that the element of the product matrix A A t i j the i jth element of A A t is since this is a an identity matrix on the right side has to be the Kronecker delta delta i j. Now, you can write this again A A t i j in terms of the elements of A and so, if we do that it is sum over k A i k and the matrix A t k j and that is equal to delta i j, but the matrix A t k j is nothing other than A j k therefore, it is sum over k A i k A j k is equal to delta i j. So, if I write to this matrix by say A 1 1, A 1 2, A 1 3, 1 3, A 2 1, A 2 2, A 2 3, A 3 1, A 3 2, A 3 3 and if I say that this is an orthogonal matrix then from this relation you see that A i k times A j k keeps the columns constant therefore, you take i k k is equal to 1 here sorry the i and j are constant the columns are summed over therefore, you have summing over elements of 2 different rows rho i which is for example, if i is 1 it is a first row if j is 2 it is a second row then A 1 1, A 2 1 times A 1 2, A 2 2 times A 1 3, A 2 3 is of course, since it is 1 and 2 is 0.  So, this is also known as row orthogonality, the rows are orthogonal to each other ok. Now, A t A is also equal to the identity and that immediately gives you the column orthogonality.  So, we take a column and we take another column if we multiply the two columns element by element and then add them all up it gives you 0 whereas, if you take the same column the column of elements A 1 1 A 2 1 A 3 1 and multiply them with themselves A 1 1 themselves A 2 1 A 3 1 and that gives you 1. Therefore, the columns are normalized to 1 the columns are orthogonal to the other columns the rows are normalized to 1 the rows are orthogonal to the other rows and this is the property of there an orthogonal matrix.    So, what you will get is let me write it here k U i k U j k star is equal to delta i j and therefore, you see this is equivalent to a rho orthogonality because you are keeping the two rows i and j constant rho orthogonality the difference between a unitary and the orthogonal matrix is that the rho orthogonality involves the complex conjugate of one of the rows it does not matter which rho you complex conjugate when you say a number is equal to another number its complex conjugate is equal to the complex conjugate of that number delta i j is 1 or 0 therefore, the complex conjugate of this number is also 1 or 0. So, it does not matter which rho you complex conjugate, but you multiply one row with the complex conjugate of another row for a unitary matrix when you do it element wise and sum them up the answer is 0 and likewise for column orthogonality you can very easily show that sum over k U k i U k j star is equal to delta i j. So, let us have a break here and right after this we will come back to looking at some properties of the Hermitian matrices and then in the last part of the lecture we shall solve the Eigen values of a 3 by 3 matrix which has degenerate Eigen values. So, we shall start with the properties of Hermitian matrices what is a Hermitian matrix? A matrix A is Hermitian if it is equal to its Hermitian adjoint a dagger which is a t star. So, what you have is a i j is equal to a j i star this is the definition of a Hermitian matrix ok. So, let us use the properties of Hermitian matrices and let us use this property to prove a few simple results result 1 Hermitian matrices have real Eigen values or let me write in this way the Eigen values of a Hermitian matrix are always real we shall see that just in a few minutes ok. Now please remember the bracket notation when we write the vectors by column and also rows if you remember that ok keep that in mind we may not need it now, but we will need it further during this part of the lecture. So, let us define the Hermitian matrix the Eigen vector 1 Eigen vector of a Hermitian matrix by 1 Eigen value lambda x ok. Now remember what this stands for? This stands for a column vector x containing elements x 1, x 2, x n and has a Hermitian matrix h i j where i n j run from 1 to n and it is equal to lambda times x 1 x 2 to x n ok. So, this statement if you have to write I can write this for one element x i the following namely sum over j h i j x j is equal to lambda times x i ok. The row that multiplies this column gives you that element in this row ok here the row here multiplied by this column gives you that row ok that is what this is h i j times x v lambda x i let us multiply both sides by x i star and then sum over i all i's i is equal to 1 to n. So, when we do that we get the following equation namely sum over i j x i star h i j x j is equal to lambda times lambda being a constant is outside the sum i x i star x i. Since it is an equation involvingtwo quantities two numbers. So, if we complex conjugate this equation both left hand side should also be equal ok. The complex conjugate of this is i j x i h i j star x j star is equal to lambda star sum over i x i x i star, but of course this is part is equal to this part, but lambda and lambda star we still do not know that is the equal point we wanted to show. Now, remember that for a Hermitian matrix h i j star. So, this theindex the indices i and j are dummy indices therefore, if we interchange i and j throughout this sum remains the same. So, if we do that i j x j h i h j i star because we are doing that interchanging x i star nothing changes on the left hand side because these two are equal and it is still equal to lambda star sum over i x i x i star ok. Now, use the Hermitian property namely h j i star is equal to h i j therefore, what we have is sum over i sum over j x i star h i j x j is equal to lambda star sum over i x i x i star, but remember this is exactly the same as the first line that we started with namely this one sum over x i star h i j x j sum over i j and this is also sum over i and sum over j x i star h i j x j, but the right hand side now has lambda times x i star x i sum over i and here it has lambda star x i x i star. Therefore, the sums are the same sum over i x i star x i and sum over i x i x i star are both the same therefore, what you have is lambda minus lambda star sum over i x i x i star is equal to 0 and this is nonzero therefore, lambda is equal to lambda star or lambda is 3 L ok. So, very simple way of verifying that the Hermitian matrices have real eigenvalues. Now, the second property is that the eigenvectors of Hermitian matrices, the eigenvectors of Hermitian matrices will be there will be different eigenvectors for different eigenvalues. One important property is that the eigenvectors of a Hermitian matrix corresponding to different eigenvalues are orthogonal to each other that is what we will next show. If the eigenvalues are the same, they are not necessarily orthogonal, but they can be made orthogonal which is the third property that we will show by defining what is known as an orthogonalization process or a Gram-Schmidt process and then we will see how eigenvectors of a Hermitian matrix with the same eigenvalues can be orthogonalized. So, these are the properties that we need to know again and again and we apply them all the time and therefore, it is important to keep this right in the beginning ok. The eigenvectors of a Hermitian matrix corresponding to different eigenvalues or orthogonal. Now, for this process we shall recall the bracket property namely h, we have h as the Hermitian matrix with one eigenvector x with an eigenvalue lambda 1 times x and we have another eigenvector of the same Hermitian matrix y with another eigenvalue lambda 2 y and we assume that lambda 1 is not equal to lambda 2. Given that we want to show that x and y are orthogonal or if you remember the bracket relations that I discussed in the previous lecture, what we want to show is the inner product x y is equal to 0 is equal to y x ok. This is what we wanted to verify. Now, the x is given by the columns here of course, it is given by the rho y is given by the column. So, what we wanted to show is sum over i x i star y i is equal to 0 or equivalently sum over i x i y i star is equal to 0. This is what we wanted to prove ok. Again we will have to use the property of the Hermitian matrices. So, let us first write h on x is equal to lambda 1 on x as sum over j h i j x j is equal to lambda 1 x i and h on y is equal to lambda 2 y as sum over k h i k y k is equal to lambda 2 times y i. Let us multiply the first equation by y i star y i star multiply by y i star sum over i this equation namely j h i j x j is equal to lambda 1 x i. So, the answer is sum over i n j y i star h i j x j is equal to sum over this is i sum over i lambda 1 y i star x i. Now, if we interchange the indices here i n j we still get the same equation because they are summed over both i n j and so what you have is sum over i n j y j star h j i x j is equal to lambda 1 sum over i y i star x i. This is h i j star because it is a Hermitian matrix. Therefore, what we have is sum over i this is y i I am sorry this is y i. So, what we have is h i j star sum over i j when we interchange the indices i n j we should have here a different index namely y j star h i j x i now it is ok. So, what we have is h i j star y j star and sum over j and we shall keep the sum over i 2 x i and that is equal to lambda 1 sum over i y i star x i. Now, this is the eigenvalue equation for the y and h i j star y j star will give you lambda 2 times y i and lambda is real. Therefore, what you have is sum over i lambda 2 you have y i star x i is equal to lambda 1 sum over i y i star x i. So, please remember that if you have the equation h i j y j star sum over j star the result is lambda 2 which is real therefore, it is a star is the same you will get y i. So, that when you substitute you get lambda 1 minus lambda 2 times sum over i y i star x i is equal to 0 and since lambda 1 is not equal to lambda 2 this quantity sum over i y i star x i is equal to 0 or this is the same as writing x the vectors y and x are orthogonal if the eigenvalues lambda 1 and lambda 2 associated with the two vectors x and y are different. So, it comes out naturally from the property of the Hermit Hermitian matrix. Now, another important property is that what if the eigenvalues are the same if lambda 1 is equal to lambda 2 lambda 1 minus lambda 2 is automatically 0. Therefore, if you have a matrix with what is known as the what are known as the degenerate eigenvalues that is two eigenvalues are the same then can we show that by this process that the two eigenvectors are orthogonal. We do not have to worry about that. If two eigenvectors are there in the first place and if they are not orthogonal it is always possible for us to orthogonalize them anyway. So, let me just to show you for a simple two eigenvectors two vectors which are orthogonal how we get the orthogonalized vector and with that in mind it is immediate that even if we do not get the orthogonal eigenvectors it is possible for us to orthogonalize them and therefore, we can assume that the eigenvectors of a Hermitian matrix are always orthogonal no matter whether the eigenvalues are the same or not because it is possible to orthogonalize them anyway ok. So, let me just show you that when two vectors x and y are such that that their inner product is not 0 that is x and y are orthogonal not orthogonal how do we get orthogonal eigenvectors out of them ok. This procedure is called Gram-Schmidt orthogonalization procedure C H M I D T and can be extended to any number of arbitrary vectors I mean vectors which are not orthogonal and it is a very simple way of visualizing what is meant by an orthogonal quantity. So, if you say that x and y are not orthogonal to each other what it means that a component of y in the direction of x is non-zero that is what it means or vice versa the component of x in the direction of y is non-zero. Therefore, if we subtract to this component from that vector what remains is obviously not going to have any common component and therefore, it is going to be an orthogonal to the original vector that is all you do you subtract out the non-orthogonal part of an eigenvector from another eigenvector through a projection process and this removal makes the resultant two eigenvectors to be orthogonal because they do not have anything in common. It is an extremely simple way physical way to visualize them and let me write the answers out very quickly ok to follow the class work notation I would denote the two non-orthogonal vectors as with the same eigenvalue H x 1 is equal to lambda times x 1 and H x 2 is equal to lambda times x 2 this is because the eigenvalue is doubly degenerate in this case that is the matrix has two eigenvalues which are equal and these are the corresponding eigenvectors and we have x 1 x 2 is not necessarily 0. Therefore, how do we make an eigenvector two eigenvectors which are orthogonal out of this first let us define the two vectors x and y as follows x is equal to x 1 and y is x 2 without the component of x 2 that is in the direction of x 1 the component of x 2 in the direction of x 1 is subtracted out and that is the scalar x 1 x 2 times the vector x 1 this is the component which when projected along the x 2 gives this part x 1 x 2 divided by the normalization quantity x 1 x 1 ok. The moment we define the vector y as the vector x 2 minus the component x 1 x 2 on x 1 which if you remember the bracket notation that we used can also be written as x 2 minus x 1 x 1 on x 2 divided by the length x 1 x 1 squared you can see that this is a projection operator and when it projects on to x 2 it gives the number x 1 times x 2 multiplied by the vector x 1 and if the vectors are orthogonal to begin with then this product is 0 and therefore,the two orthogonal vectors are x 1 and x 2 themselves which is equal to x and y, but if this product is non-zero then you only remove that magnitude from x 2 and then what you get as y will be now orthogonal to the vector x. So, if we take the product x y you can see that x is equal to x 1 so it is x 1 x 2 minus x 1 x 1 times x 1 x 2 divided by x 1 x 1 and you know that this goes away and so what is left over is that x 1 and x 2 this cancels out this is 0 therefore, you see that x y is orthogonal ok. Now or x and y the same eigenvec values they do they have the same eigenvalues it is easy because h on x is the same as h on x 1 that is how we defined x and therefore, it gives you lambda times x 1, but what about h on y that is also easy to see because h on y is h on x 2 minus you remember it is x 2 x 1 times h on x 1 divided by x 1 x 1 and this is of course, this is lambda times x 2 minus lambda times x 2 x 1 on x 1 divided by x 1 x 1. So, you have lambda times whatever is in here is the vector y therefore, h on y is equal to lambda times y. So, y is the eigenvector with the same eigenvalue except to that now y and x are orthogonal. So, this process is called the Gram-Schmidt orthogonalization. Now what about if we have three vectors which are not orthogonal, but all of which have the same eigenvalues. So, if we have a larger Hamiltonian matrix at 4 by 4, 10 by 10, 20 by 20 and if some eigenvalues are the same how do we find for such Hermitian matrices the orthogonal components same process. From each vector you remove all the components that this vector will produce when it is projected to all those vectors you subtract them out then you will get the eigenvectors which are orthogonal you will have to order them. So, for example, you will have to choose one vector as a starting point as the first vector you will have to choose x 2 as a next vector any one of them sorry this is x you will have to choose the second vector y as x 2 minus the component of x 2 in the direction of x 1 and then you will have to choose z as the component of x 3 in the directions of both x 1 and x 2 subtracted out from x 3 if you do that then obviously, z will be orthogonal to both y and x and therefore, you can create a chain of orthogonal eigenvectors in a sequence with some other sequence of non orthogonal eigenvectors. I will give that as a problem in one of the assignments, but it is important to recall that these things are fundamentally important. Now, let me pass for a break and in the next 10 minutes I shall solve the eigenvectors of a 3 by 3 matrix as an example and then we will stop here we will continue with the properties of matrices in the next lecture. So, the third section will be eigenvalues of a 3 by 3 matrix. So, we shall look at the eigenvalues of a Hermitian matrix again with the degenerate eigenvalues. So, I shall use a well known textbook example namely a matrix A 502010202, I can use any other matrix I found that this is there in the book of Arfkian and just I will solve this. This is also referenced in my lecture notes I use follow the book of George B. Arfkian and his colleague Weber the mathematical methods for physicists it is one of the best books on mathematics at the graduate student level that I have come across. I use it a lot I am thankful to the authors for having written this through several editions and now I believe the book is on the sixth edition at least that is what I have as the latest one. And some examples occasionally I shall pick up from there even though I will try and solve mostly problems of my own this is an example picked up from that book ok. Now very simple the eigenvalues of this matrix eigenvalues ok one can obtain this by the determinant of phi minus lambda 0 to 0 1 minus lambda 0 to 0 2 minus lambda that is equal to 0. And if you expand to this immediately you get phi minus lambda times 1 minus lambda times 2 minus lambda and if you expand it along either the row or the column it does not matter plus 2 times minus 2 times 1 minus lambda is equal to 0. So, what you have is 1 minus lambda times phi minus lambda times 2 minus lambda minus 4 is equal to 0 the answer is lambda is equal to 1 this is 7 10 minus 4. So, it gives you 6. So, the equation here is lambda square minus 7 lambda plus 6 and then you have 1 minus lambda and if you factor this out it is 1 minus lambda times you factorize that you get lambda minus 6 times lambda minus 1 is equal to 0. So, you have lambda is equal to 1 as twice the eigen 2 eigenvalues and lambda is equal to 6 as the other eigenvalue. So, here is the degeneracy ok. Now we shall use the fact that the eigenvalues of a Hermitian matrix can always be eigenvectors of a Hermitian matrix can always be orthogonalized and therefore, we shall assume that the eigenvectors would be orthogonal we will use that property ok. So, here lambda is equal to 6 the eigenvector is easy to obtain you have phi 0 2 0 1 0 2 0 2 you have this is lambda 1 then you have x 1 1 x 2 1 x 3 1 is equal to 6 times x 1 1 x 2 1 x 3 1. So, the equations that you get are minus minus x 1 1 plus 2 x 3 1 is equal to 0 and here you get anyway minus 5 x 2 1 is equal to 0 and these the third equation that you get is obviously, going to be the same thing namely minus 2 x 1 1 plus 4 x 3 1 is equal to 0 these 2 are 1 and the same. So, you have got the equation that x 1 1 is equal to 2 x 3 1 and x 2 1 is equal to 0. So, the eigenvector not normalized is 2 a 0 a and if you normalize it you will cancel the a out and you will get 2 by rho 5 0 1 by rho 5 this is one eigenvector. Now, the eigenvector 2 for eigenvalue 1. So, when you write that you will see the redundancy right away 5 0 2 0 1 0 2 0 2 x 1 2 x 2 2. So, this is x 3 2 is equal to 1 times x 1 2 x 2 2 x 3 2, but since 1 happens twice there are 2 such columns that we have to find out. So, let us call this one as lambda 2 is equal to 1 ok that is why this 2 and you can see immediately that it is 4 x 1 2 plus 2 x 3 2 is equal to 0 and you have x 2 2 minus x 2 2. So, it is there is no other equation because this x 2 2 will cancel out with this x 2 2. So, x 2 2 is arbitrary in addition to or undetermined in addition to 1 of the 2 being arbitrary that is what it is when you have a degenerate eigenvalue doubly degenerate eigenvalue. If you have a triply degenerate eigenvalue you will see that 2 of the eigenvector coefficients will become will not be involved in any equation in addition to the third one. So, you will see that these things are not without any structure I mean there is a fundamental structure the eigenvalues reveal that when you look for the symmetry that is degeneracy refers to some hidden symmetry in the system and here the question is we do not know anything about x 2 2. We know that x 1 2 has to be half of minus the half of x 3 2 ok. So, we have one equation we can choose x 2 to be either 0 or x 2 2 to be a constant ok. So, both choices are possible x 1 2 is of course is equal to minus x 3 2 by 2. Therefore, if we choose x 2 2 is equal to 0 then the eigenvector that we have is 1 by root phi 0 minus 2 by root phi or if I put the minus here plus here. What about choosing x 2 2 is equal to a? If we choose x 2 2 is equal to a or some constant then and choose x 1 2 to 0 then what we have is 0 a 0. Now, this choice is meaningful in the sense that these two eigenvectors are orthogonal to each other. You can see that the column this multiplied by that this multiplied by that this multiplied by that the sum of that is 0. So, the choice of x 2 2 is equal to a x 1 2 is equal to 0 gives you immediately an orthogonal eigenvector to the other eigenvector both of which have the same eigenvalue. So, it is possible to do that, but a is of course, when you normalize it becomes 0 1 0 ok. So, you have 3 eigenvectors lambda is equal to 6 corresponds to 2 by root phi 0 1 by root phi and lambda is equal to 1 corresponds to 2 eigenvectors 1 by root phi 0 minus 2 by root phi and lambda is equal to 1 corresponds to 0 1 0. Now, if you do not choose this 0, but you choose it to some arbitrary you have to ensure that this eigenvector is anyway orthogonal to this eigenvector ok. So, you can see immediately that if x 2 x 1 2 is equal to x 2 x 1 2 is equal to minus x 3 2 if you have that you have to ensure that this quantity and this quantity these both eigenvectors are orthogonal. So, I made the easier choice of making them to 0, but you can play around the point is when eigenvector values are degenerate you have to be careful with the way you derive the eigenvectors. It is always better to do the eigenvectors one at a time and make sure that the second eigenvector is orthogonal to the first one and also to other eigenvectors and so on. So, it is a it is a process that one has to do carefully of course, you will learn this as you go along and as you solve more and more problems ok. So, let me stop at this point the property or the characteristics of eigenvalues and eigenvectors and in the next lecture for the first part I will talk a little bit about the functions of matrices and matrix operators as we call them or functions of the operators we will talk a little bit about the commutativity and properties of matrices and then the second half of the lecture we will move on to the next topic namely the solutions of differential equations. Now it is not possible for me to talk a lot about matrices in this course except to the point up to the point that we need them and we use them as we need more and more such matrix techniques we will develop them then and there, but it is important to take a look at the differential equations first order and second order again in a rather quick way in order to start the process of solving the Schrodinger equation which we will start off as a differential equation and later on only we will use it as an eigenvalue equation for matrices. So, until then thank you very much.


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But before that, if it's the first time of you coming to my channel, remember to hit the subscribe button as well as the notification bell so that you will not miss out any of my future investment updates. An early thumbs up is also appreciated because I will tell YouTube algorithm that you find this video helpful, so that more people will get to see this video to learn how to invest safely. In return, I will share with you how to invest in banks in this volatile crisis. Before you invest in banks, there are several metrics for you to consider to assess the bank's financial health as well as future growth potential. That's why in this video, I will use Bank of America as well as the recent fell SVV bank as an example, so that you learn how to look into them and analyze for yourself. So here are the top 5 investing metrics to consider before buying your first bank stock. 1. Non-performing Loan to Total Loan Ratio In order for banks to make money, they need to loan out their money to earn an interest. However, if the borrowers are unable to pay back, it will eventually snowball into a huge problem just like what we saw back in the old 8 financial crisis. That's why the first metric to check is the Non-performing Loan to Total Loan Ratio. This metric measures the percentage of the loan not being paid back on time, and a high MPL ratio represents that this bank has a higher default risk, and therefore is more financially unstable. That is why we want this ratio to be as low as possible with a maximum of 2% to be safe. By looking at SVV's MPL ratio, it was a very healthy 0.18% in the most recent annual report. And for BAC, it was 1.22% for year 2022, which is also within the reasonable range. So based on this metric, SVV shouldn't have failed. Exactly, that's why we need to dig deeper to look into investing metrics number 2. Net charge-off rate percentage If there's still a chance for the bank to collect back its Non-performing Loan, there will be no chance at all when it comes to net charge-off. Also known as bad debt, net charge-off are loans that deem to be irrecoverable, and that's why the banks return it off as an expense. If you look at BAC, it has a net charge-off rate of 0.21%. In fact, inside my private telegram, I share with my followers that BAC has been enjoying a low net charge-off rate since the await crisis, be it in commercial, consumer, or credit card segment. So that means since the await crisis, the bank has been becoming really cautious in terms of lending. If you haven't joined my telegram yet, make sure to join by the link below to get the latest update, and it's free. Now let's take a look at the net charge-off rate of SVB. Oh my god, it's even lower than BAC at 0.1%. So why did it collapse? So it shows that just by looking at these two metrics, it is not enough. That's why we need to look at the next one, which is investing criteria number 3, growth in deposits. In order for banks to loan out money to an interest income, they need money in the first place as a form of deposit. That's why if the bank has a consistent growth in deposit, this is a good sign. On the other hand, if the bank has been consistently having less and less deposit, this could be a dangerous sign. So let's take a look at SVB. As you can see, SVB has enjoyed a 25% growth in deposit as compared to 2021. What for BAC deposit decreased by 6.5% as compared to the year earlier? You must be wondering, if that's the case, SVB looks like a better bet to invest in, isn't it? Before jumping to the conclusion, let's take a look at the next metrics, which really marks a huge difference between BAC and SVB, which is size of the deposit. Although SVB enjoys a 25% growth in deposit, its size is only $185 billion, as compared to BAC's $1.9 trillion. The sheer size of the deposit just makes BAC a much safer bank to invest in, because most of the deposit sources come from normal people living in the US. Think about it, how likely would it be for Americans to want to withdraw cash from the bank and put it as cash inside their house? The thought of it sounds even more scary, isn't it? That's like inviting robbers to come to your house. That's why it's very unlikely for BAC to suffer bank run, especially with the government's assistance to ensure the stability of the financial sector. So people in the US will definitely have more confidence in big banks like BAC, JP Morgan, rather than small regional banks like SVB. On the other hand, for SVBs, most of the deposits come from tech startups, venture capital firms and private equity, which are a lot less stable in nature. So when all these companies start to smell trouble, they quickly withdraw funds from SVB, and thus causing the liquidity crunch which eventually causing the bank to collapse. Of course, analyzing banks takes more than just looking at these four metrics. Investors should also look at other criteria like return on asset, loan portfolio, net interest margin, and so many more. Just like what Warren Buffer said, it's actually a lot more difficult to understand a company from the financial sector. Financial companies are more difficult to analyze than many companies. If you're analyzing something like WD-40 or C's candy or our brick business, whatever, they may have good or bad prospects, but you're not likely to be fooling yourself much about what's going on currently. But with financial institutions, it's much tougher. So if you don't understand individual banks well enough, here comes the two number five to help you to invest in the financial sector safely, which is to ETF. By buying an ETF like XLF, you immediately build a portfolio with all these strong companies inside, including Warren Buffer's Berkshire Hathaway, JP Morgan, Bank of America, Amex, and more. And for the past 10 years, the ETF has generated investors close to 10% return. Pretty amazing, isn't it? The current stock price is $31, but if you know how to use BOSS option strategy, you can buy XLF even cheaper than the current market price. For example, true BOSS option strategy, by promising to buy XLF at $30, you get to collect about $67 in terms of passive income in 30 days. That's a 2.2% return in one single month. Of course, there are different option strategies for you to take advantage of in different market conditions. If you want to learn how to get started, your stock and options investing journey so that you can collect passive income consistently even in this volatile market, do join us in our upcoming 3-2-hour next-level options masterclass where we'll share with you 3 different option strategies for you to get started step by step. All you have to do is click on the link below and register for your free spot. Of course, the collapse of SVB also has to do with other factors such as the company buying a lot of long-term treasury bonds thus causing the liquidity issue. It just shows that blood swan event can happen anytime, and we should never take our investment portfolio for granted. That's why it's so important that we invest in something that we truly understand, and when it comes to investing, it's very important to manage our risk. And that's why regardless how confident you are in one single company, it's always important to build a portfolio so that in case this company doesn't do well, your portfolio is still intact. And if you're not sure about one particular company, it's always good to invest with ETF so that you can immediately build a portfolio for yourself. So what's your greatest takeaway from this financial crisis? Comment down below and let me know your thoughts. In the meantime, don't forget to join my Telegram channel so that you can get more investment updates. Also check out some of the videos right here to continue your investment journey. With that, I will see you in the next video. Arigato!


He's coming up on me now quick. He's just going round me on double lines in the 60s zone. He's doing 73 in the 60s zone He's got a full mile premium on the back. Okay, overtook me on double lines that got up to 72-73 in the 60s zone overtook me near the cafe there Okay, I'm not behind it. We've got a VT Commodore between us. He's about to go around that too.  Okay, he's just overtaking the Commodore


Let's predict the Browns record this season the Sean Watson suspended for six games So that's what we're gonna go with on this prediction the first game against the Panthers I think Baker gets his revenge and I also think they lose their home opener against the much improved jets now They come home and they they get a win against the Steelers in prime time I like him to win that game and then they alternate wins so Falcons is a win chargers is a loss Patriots is a win Ravens is a loss Bangles on Monday night football is also when heading into the bot Coming out of the buy we got the Miami Dolphins I think they can win that game fresh off the buy, but then they lose two in a row against the Bills and the Buccaneers You know when went on the road against the Texans I think they take the L on the road against the Bangles But then they go on a four game winning streak to end the season against the Ravens the Saints and some tough games And a little bit easier with the commanders the Steelers will probably have Kenny Pickett starting So I give them a win there as well. Let me know your guys's thoughts on the Browns record this season


Hi there, my name is Jean-Clapper and I am a surgeon. One of my passions is medical education. In this series of presentations, I'll talk about things surgical. Remember, these presentations are available for continuous professional development points, so I leave the description to that down below. In this presentation, I'm going to talk about thyroid nodules. As we do more and more ultrasounds, we discover more and more thyroid nodules. In actual facts, some reports state that up to 70% of all patients will develop thyroid nodules. So in this presentation, I'm going to tell you what to do with this very common diagnosis. The nodules of the thyroid are amongst the most common histological disorders in the human body. Actually more than 70%, 70% of all humans will develop a thyroid nodule or even more than one nodule during their lifetime. This number keeps increasing with ever more sensitive imaging techniques and the incidence increases with age and there is definitely this female predominance. Most of these nodules of the thyroid are fortunately benign. This is so in at least 90% of cases and most patients will never ever be aware of them. Even some small cancers remain subclinical. Now fear response is triggered in most patients in whom a nodule is detected and it's our duty, the duty of the clinician to manage such a finding with great care. With nodules being so common and routinely seen on ultrasound imaging, it is important to know how to manage the patients with a thyroid nodule. We have to look at the different types of nodules. Now thyroid nodules may be colloid filled cystic lesions all the way over on the other side of the spectrum we find neoplasms. Now the latter are either benign, those are true monoclonal adenomas or follicular adenomas or they can be malignant. Now a combination of cystic and solid nodules are commonly seen and that happens in multinodular goiters. A nodule can be cystic to a varying degree from completely cystic, that's a true cyst with an epithelial lining but that's quite rare and then also over to only containing a small cystic component. Cysts are usually the results of a degenerative process of the solid component of a nodule. It is a rule of thumb that the malignant potential of a nodule decreases as its cystic component increases. Now that's relative to the solid component in the nodule. True monoclonal adenomas are the most common solid lesions of the thyroid. These can usually be identified on an ultrasound and on aspiration cytological examination and they are really benign lesions. Follicular new plasms, they can either be benign or malignant. Now cells seen on cytology for both benign and malignant follicular lesions appear similar and the diagnosis can only be made on histological evaluation of the surgically resected specimen. There are various types of thyroid carcinomas. These include papillary, follicular, hurdle cell, medullary and aplastic carcinomas as well as primary thyroid lymphomas. Yes, you can also develop a lymphoma in the thyroid. Now the true proportion of malignant lesions they really not known as it depends on how the incidence of nodules as a whole is described. If we only look at the incidence of clinically discovered nodules, then the incidence of carcinoma is going to be high. But of all true nodules of the thyroid, the incidence is pretty low. Let's look at the risk factors for the development of thyroid nodules. The most common risk factors are advancing age, the female gender, dietary iodine deficiency and a history of exposure to ionizing radiations such as radiotherapy of the head and neck. So the most common ones there really age and the female gender. Now these are not the only factors that put a patient at risk. The tobacco smoking, excessive consumption of alcohol and obesity are risk factors for many diseases including unfortunately also the development of thyroid nodules. Now the less commonly known risk factors are the presence of uterine fibroids, high insulin growth factor 1 levels and hematopoietic stem cell transplantation. So some of the rare risk factors there. Now thyroid nodules and cancers can be associated with familial disorders and it is important to inquire about a self or a family history of thyroid disease or cancer or indeed any endocrine conditions. Remember multiple endocrine neoplasias that's especially types 2A and B they are associated with medullary thyroid cancer. Now familial non-medullary thyroid cancer that's quite rare and it's really not that well defined. It is known to be associated with cardin syndrome, familial adenomatous polyposis, gardener syndrome, carny complex type 1, venous syndrome and DISA 1 syndrome. So all these are very rare conditions. Now some factors are associated with a decreased incidence of thyroid nodules so something that decreases the patient's risk of developing a thyroid nodule. Now these include the use of the oral contraceptive pool and also statins. Now unfortunately it's not currently known if these hormone formulations and medications are protective in any way though. We just know that it is associated with a lower incidence of thyroid nodules. Now size of the actual lesion that is a risk factor for malignancy and the larger they are especially after about two centimeters in greater diameter it becomes more likely to harbour a carcinoma. Clinical findings now depending on the clinical setting thyroid nodules are either diagnosed when they become visually evident so that's visually or through their symptoms or when seen incidentally during imaging for other conditions and most notably the common use of ultrasound these days. Now whilst a visible neck mass is common in some settings and yet others these nodules are palpable only so you don't see them by looking at the patient but can palpate them. Now some that are large enough especially those in the periphery of the gland can cause compressive symptoms involving the aerodigestive tract. Now these compressive symptoms are dysphagia, a dynaphagia, choking, this feeling of a squeezing sensation in the neck, a globus sensation that is the sensation of a mass or a lump or a foreign body in the throat that doesn't disappear by swallowing and then also dyspnea and some voice changes that subtle voice fatigue in cases of large nodules or actual voice changes due to nerve infiltration by a malignant process. Now the thyroid gland is visually inspected from in front of the patient, stay in front of the patient and we make note of any gross enlargements or asymmetry and then the thyroid moves during swallowing and remember all signs should be elicited by asking the patient to swallow. It's difficult when someone just asks you to swallow so it might be handy just to have a glass of water nearby. Now breweries and thralls they are not commonly seen in a thyroid nodules and they really may be indicative of other pathologies such as carotid aneurysms so watch out for those. Now the thyroid is palpated from behind the patient, once again asking the patient to swallow and then a note should be taken of any asymmetry and also the surface of the thyroid as well as its size, consistency, location and the movement of any masses. Looking on the patient's habitus and the clinician's experience with examination of the thyroid the finer changes may be missed, we all missed some of these, the finer changes that you find with these, the appearance of these nodules. Now at the very least we can take note of the following rule of thumb, bilateral enlargement with a smooth surface, smooth surface that's indicative of a goiter that's iodine deficiency or even Graves disease, a unilateral enlargement is indicative of a nodule so just feel that one side is bigger than the other that's probably a nodule and then an irregular surface and a normal size or an enlarged thyroid that's indicative of a multi nodule of goiter so that's just this rule of thumb and then nodules that are rock hard and immobile that means they fix to the surrounding structures that should raise the suspicion of a malignancy. Now malignancy can metastasize to original lymph nodes, we think about papillary thyroid carcinomas and also the jugular chain and we should really palpate these to feel for nodes. Now the famous Pemberton sign that you might remember that involves venous jugular engorgement, facial plethora and even dyspnea when the patient lifts their arms that's the Pemberton sign, we don't see that commonly but it's worthwhile to remember now that's caused by severe compressive symptoms due to a large retro sternal goiter. What about thyroid function test we all send off our patients off to the lab to get to get thyroid function test now we really have to make a special note of this most nodules are hormonally inactive but they can overproduce thyroid hormone in some cases and yet others they occur in the setting of hyperthyroidism such as that caused by Graves disease or in the setting of hypothyroidism such as caused by Hashimoto's thyroiditis so we really have to be a look in the lookout on the lookout for thyroid nodules in all of these on these cases now thyroid nodule in a hyper thyroid patient that necessitates the use of a centigram and that's to determine whether it's the nodule causing the overproduction or the underlying thyroid itself so yeah send away for your thyroid function tests so how do we work up these patients well the absolute minimum workup after discovering any thyroid nodule includes a thyroid stimulating hormone level that's number one and then a thyroid ultrasound and then further work up is determined by the results of these two investigations so thyroid function test and an ultrasound. Logistic regression analysis through research that's shown that patients with a normal TSH level those with the level just close to the upper end it's about 1.7 million inspilita they have higher odds ratio of malignancy than those that are still normal but at the lower end of normal so just something to remember so in the case of a suppressed TSH level that's hyperthyroidism overactive thyroid we do a thyroid syndrome with radioactive iodine isotope now the possible reporting will either be that of a hot nodule or a warm nodule or a cold nodule and the warm and cold ones they follow up as the same as someone with a normal TSH level so those with a hot nodule we don't really do any further work up for them because carcinoma just at least for investigating the possibility of a cancer it's very rare to have a hyperfunctional thyroid carcinoma. Now these patients with their hot nodules they are further evaluated for radioactive iodine therapy or surgery and then this includes the pretreatment of an antithyroid medication to manage their hyperthyroidism and unless there's a contraindication to surgery a thyroid or lebectomy and isthmus resection in the case of unilateral disease or total thyroidectomy in the case of bilateral disease that has a very high cure rate for this overactive nodule. Now hyperthyroidism they that occurs then in less than 3% of cases following a lebectomy and unilateral disease but in the case of radioactive iodine therapy if the patient chooses that that has a higher recurrence rate about 1 in 5 will have recurrence of their hypothyroidism. Now up to a half of patients they will also develop hypothyroidism within 10 years after the therapy so we have to follow these patients up long term. So then let's look at the patients that are euthyroid or hypothyroid. So they now go for the work up because we need to determine whether this nodule is malignant or not. Now the findings on ultrasound, ultrasound is our first investigation. The findings on ultrasound are classified by the American College of Radiologists using the Tyreds system and you get this kind of system now for many possible lesions in the body. Now the aim of the imaging classification system that's to reduce the incidence of biopsies on lesions that are highly unlikely to be of clinical significance. So we have the classification system from our radiology colleagues because we really don't want to to overdo things just to take by unnecessary biopsies and just increase the cost and the stress to the patient. We have the Tyred system here we look at a few the radiologists will look at a few things the composition the ecogenicity the transverse plane shape the margins and the ecogenic foci in this lesion and points our given for those those are then tallied and we get we get the result. What we have is a Tyreds 1 2 3 4 5 and a Tyreds 1 that's really a benign. 2 is a non suspicious lesion 3 is mildly suspicious 4 is moderately suspicious and 5 is then highly suspicious. So for benign we don't do any further work up there's no fine needle aspiration cytology required and also for a Tyreds 2 for a mildly suspicious lesion that's type 3 we can really follow that patient up for a year a year 1 year 3 and year 5 and that is if the lesion is less than 25 millimeters in greatest diameter if it's more than 25 millimeters in greater diameter we do a fine needle aspiration cytology. When we get to moderately suspicious that's Tyreds 4 we also follow the patient up on years 1 3 and 5 and that's for lesions less than 15 millimeter they 15 millimeter more fine needle aspiration cytology. When it comes to Tyreds 5 if the lesion is less than a centimeter in greatest diameter we can do a follow-up of these lesions they small and we're not sure of their clinical significance and if they are greater than a centimeter though they get fine needle aspiration as well so that's how we use the Tyreds system so let's look at the fine needle aspiration cytology that is performed under ultrasound guidance and we use a small 25 or 22 or 27 gauge needle and that's inserted right into the lesion and then the refinings are reported by our pathology colleagues using the Bethesda system for reporting thyroid cytopathology and there are six classes described using this Bethesda system now a class one that the class identify samples that are not amenable to psychological examination we see that put the needle in and it's unfortunate bloody aspirate and we can't make a diagnosis on that there's this inadequate sampling it is also seen when lesions are particularly cystic sometimes it's impossible to obtain a good sample in the case of the latter clinical judgment must be used to decide between a lobectomy and interval observation and we usually do that according to the Tyreds then into according to the size now most cases a repeat aspiration will need to be done and it usually yields a proper sample now it's best to do this in the setting then where the aspirate can immediately be examined in the presence of a pathologist where they can look for adequate follicular cells so sit by the patient and we take those final aspiration cytology have it looked at immediately not to make the diagnosis but just to make sure that there are follicular cells that can be examined now remember this advice at three months past before repeating the final aspiration so you can't just do it the next week unless there's a very high suspicion of malignancy then we then we can do it sooner but it is advised to to have three months in between those final aspirations for class two that's clearly benign pathology in an adequate sample and that's the most common diagnosis then is that of a benign colloid nodule with benign appearing follicular cells and an abundance of colloid so this colloid cell ratio and the more colloid with adequate cells that really tells us that it's benign now care should be taken to ensure that the sample was representative that it wasn't taken outside of the nodule because there's a false negative rate and then the American Thyroid Association they advised that all patients should have a follow-up ultrasound within the following year to document growth even if it is this Bethesda 2 so growth we define that as an increase of 20% in at least two dimensions so the repeat aspiration for cytology should be done in all cases of growth as well as in cases with the new compressive symptoms or visual enlargement or appearance of new nodules then we're gonna do it in FNA and again again if there's no growth identified and we can just repeat the ultrasound examination with an interval of about two years it must be remember though that benign lesions can actually grow too so you might repeat those biopsies and it is again it's again benign even even though they've grown so for class 3 that's either cellular atypia or a follicular lesion with undetermined significance so those are both class 3 now the incidence of carcinoma in this class is unknown since most of these cases do not come to soji we don't operate them and the true rate of malignancy is therefore unknown it repeat aspiration for cytology is one possible choice now if the class diagnosis persists then we'll do a lobectomy which should really consider lobectomy an isthmus resection then and in advanced areas we get molecular molecular testing these days it's becoming more widely available and if it is we'll do that for classes 2 to 5 so if we test the 4 that is a follicular neoplasm that the sheets of follicular cells that's seen with with a paucity of colloid so there's many more cells as a cellular colloid ratio that goes in the favor of being much more cellular now the malignant potential of a follicular neoplasm we can't see that on cytology because the architecture the cellular architecture with the tissue architecture should say is not preserved with a with a final aspiration now we really need that architecture the tissue architecture in the evaluation because we need to look for infiltration to make this a follicular carcinoma and how to manage these well exactly the same as class 3 we get to class 5 as really suspicious of the malignancy and surgical resection is really advised now for lesions less than 4 centimeters without lymph node enlargement or any high grade pathological findings we can do just the lobectomy an isthmus resection but in all other cases will do a total thyroidectomy 6 then it's very clearly malignant and these include papillary carcinomas it's you can diagnose papillary thyroid cancer on a final aspiration you don't need you don't need the tissue architecture now it must be noted that surgery can proceed irrespective of a low but test a class patients with benign disease may not be able to live with the compressive symptoms or they may prefer surgery for cosmetic reasons reasons that's important to discuss the possible complications of subtle or more profound voice changes due to nerve injury and the risk of hypothyroidism and with all the patients so that they can make an informed decision that is best for them it is just as important to consider patient comorbidities both current and future and it might be proven to operate before comorbidities naturally worsen with age and a higher incidence of developing with age so we really got to take all these things in consideration and discuss these with the patients so they can make decision that is really best for them now the guidelines are really unclear as to how long to follow a patient with benign final aspiration cytology and then many are seen yearly with repeat ultrasound and aspiration biopsies and they might just become sick of the situation and they might decide on on surgery as well now what to do with these cystic lesions now as we've mentioned some modules are predominantly cystic in nature most of them form from a degenerating adenomas they just degenerate and the majority are benign although malignant cystic lesions they do exist are a but they do exist now purely cystic lesions do not require final aspiration you see an ultrasound can clearly see that this is a cystic lesion and don't really need a final aspiration cytology of that now you can aspirate the cyst to dryness if they are large and they come have compressive symptoms the patient has compressive symptoms now the aspirate this we do send that for cytological evaluation and the results are managed accordingly assist that recur after repeated aspiration or remain present after aspiration especially if they symptomatic pain or compressive symptoms that can be managed by surgical lobectomy fortunately there's not many of those patients but we really can't consider the surgeon setting now the same can be said for cysts that are a diagnostic dilemma we get these mixed solid cystic nodules and in those cases they just manage them as if they are just purely solid nodules so in conclusion the entire nodules are commonly seen in all our medical practices now patients should be investigated number one for hyperthyroidism that's the number one so you send away your TSH if that's low that's hyperthyroid patient and they should undergo a scintigraphy if it's a hot nodule they require lobectomy and if it's warm or cold nodule they just treat it the same as the euthyroid or hyperthyroid patients and those that are then euthyroid or hyperthyroid they get an ultrasound and based on that they'll get a final aspiration cytology and then we decide between surveillance and surgery and then the highly suspicious ultrasound or cytology classes they are managed by surgical resection and careful follow-up it's really planned in all other cases so your thyroid patient really stays with you and you do follow them up over a long period of time as with most thyroid diseases the thyroid or pathology of the thyroid changes with time and we need to to monitor those patients please remember that this video series is part of a program of continued medical education and the information for that will also be in the description down below