Welcome back everyone. In the previous video, we actually saw an example of how one can calculate a matrix representation for a linear transformation relative to any basis of the domain and co-domain Vs and W here. But it turns out that if you choose different basis for V and W, because there are lots of different bases one could choose for these two different vector spaces, choosing different bases will give you a different matrix representation. So how much do these matrix representation change and how much do they stay the same? And up to a point, they do actually kind of stay the same. It does depend on the basis chosen, but only up to a point. And so this theorem right here is gonna give us some explanation of what we mean by that. So imagine we have a linear transformation T which goes from V to W. And we have two different matrix representations. You have A and A prime. These are two different representations of the same linear transformation T. Well, let's assume that W is m dimensional and that V is n dimensional.  Then there's an n by n dimensional, n by n non-singular matrix Q such that A equals A or P, A prime Q inverse. So there's some way of factoring one of the matrix representations into the other one using these non-singular matrices. And since this is non-singular, of course, you could rewrite this equation in the following way. We get that AQ is equal to P, A inverse, like so. And these, again, aren't any non-singular matrices. They have an inverse of some kind. And the basic idea behind the proof of this right here is that you're gonna take, these two different representations came from two different choices of bases. So let's assume that A, for example, A is the matrix representation of T where the domain has B coordinates and the co-domain has C coordinates right here. And let's assume that A prime is formed by taking T right here and you take B prime coordinates and C prime coordinates. So you have these two different coordinate systems right here. Well, then the matrix P, the matrix Q, let's start with Q, actually. Q is just gonna be the change of basis matrix where you go from B prime coordinates to B coordinates. And then the matrix P is just gonna be the change of basis matrix where you go from C prime coordinates to C coordinates. And so we can change from one matrix representation to another by some appropriate change of coordinates and change of basis right here. That's how these things are connected. So there's a really important special case of this that I wanna make mention right now that if your linear transformation is a transformation from a vector space back to itself, what if the start and stop are the same space V? Well, if you're starting and stopping in the same space V, then you actually could choose the same basis for both the domain and the co-domain. Let's say that basis is B. Then we can talk about the matrix representation relative to a single base, a single basis, and we'll denote that right here. A equals the, is the matrix representation of T relative to B. And this will then have the property right here that A, if you multiply the vector in coordinates, you'll get the output in those same coordinates as well. And so a really critical example of this phenomenon is actually the derivative, right? So we're familiar with the derivative from calculus one. We've probably seen this before here. And so let's actually construct some vector spaces with respect to this derivative. Let's take for example, some polynomial space. So we'll focus on cubic polynomials for a moment. So if we take the vector space that's spanned by the monomials one X, square and X cubed, be aware, this is the set of all polynomials of the form, A0 plus A1X plus A2X squared plus A3X cubed. And then of course, A0, A1, A2, A3 and A3 this last one. These are all just real numbers, something like that. So we can think of this polynomial space as a vector space. And we can represent vectors. If we take this generic element right here, this generic element, this generic cubic polynomial, we can identify it with the column vector A0, A1, A2 and A3. So this is a way of putting the vector into B coordinates, where B is this spanning set right here, this basis. Okay, and so if we were to consider what does the derivative look like? What's the derivative when we take all these different, if you take the derivative of all these different things, well, the derivative of constant is zero. The derivative of X here, whoops, derivative of X here is just one. The derivative of X squared is 2X. And then finally, the derivative of X cubed is equal to 3X squared. And so if we think about this in coordinates, in the coordinate system here, we start off with the vector 1, 0, 0, 0. And when we take the derivative, let's see, we'll write d, dx, and we'll put this in coordinates here. If we take the derivative of that, we end up with just the zero vector. If we do the next one, we take this in coordinates, the vector is 0, 1, 0, 0, this maps to the vector 1, 0, 0, 0. Because when you take the derivative of X, which is the second vector in our basis, you'll get one, which is the first vector in this basis. Looking at the derivative of X squared, this in coordinates, X squared looks like 0, 0, 1, 0. But when you take its derivative, this will become the vector 0, 2, 0, 0. The derivative of X squared is 2X, and 2X is two times the second basis element here. And then lastly, if you take the derivative of X cubed in coordinates, X cubed is the vector 0, 0, 0, 1. When you take the derivative, you end up with 0, 0, 3, 0, right here. And so then putting this all into a matrix, we could then say that the derivative, the derivative here with respect to B coordinates, this will be the four by four matrix, which we get 0, 0, 0, 0, 0, 1, 0, 0. I'm sorry, that's the wrong one. We wanted to do 1, 0, 0, 0. Next we get 0, 2, 0, 0. And then finally we get 0, 0, 3, 0. And so the derivative has a matrix, right? The derivative is in fact a matrix. It's this upper triangular matrix. It's actually strictly upper triangular. It's nilpotent. It's eigenvalues are all zero, right? Your eigenvalues are in fact 0, 0, 0, 0, 0. Zero shows up four times along the diagonal right there. And it turns out that this analysis of the matrix says something about the derivative map. Because after all, the derivative is a linear operation. So we can represent it as a matrix. Let's look at one more example here. Let's take the space w to be the span, the span of sine of x and cosine of x, right? Well, what's the derivative of sine? And we'll call these things C coordinates this time. Well, we know from basic calculus that if you take sine of x, its derivative is equal to cosine of x. And if you take the derivative of cosine of x, you end up with negative sine of x. And so if we wanna think of the derivative in terms of C coordinates, so if we think of it in terms of just sine and cosine, the derivative of sine is cosine. That's gonna give you zero, one. And the derivative of cosine is negative sine, which gives you a negative one, zero like this. And so the derivative map here is it's kinda curious here. The derivative is a rotation matrix. If you take the derivative of sine and cosine, you end up with a rotation matrix, which I always get this confused. This is 90 degrees rotation. It's 90 degrees counterclockwise. If I fudge there, sorry about that. And that's just kinda always blew my mind that the matrix representation of the derivative of trigonometric functions gives you a rotation matrix. This isn't some coincidence of any kind. It says that this derivative is a linear operation and we can represent matrices using, we can represent these linear transformations as matrices, which is really a cool feature right here. And so a corollary to the theorem that we had solved previously is that if you take a linear transformation from V into V, well, then you can use the exact same basis for the spaces V and V here. And so if you have two different matrices A and B, two different matrices, then there's gonna be a single non-singular matrix P so that A factors as P B P inverse. And this is just a special case of the previous observation where you can actually set P equal to Q from that previous theorem. And particularly this shows you that if you look at matrix representations where you can use the same basis for the domain and co-domain, then the matrix representations have to be similar with each other. And so because they're similar, anything that's invariant over a similarity class must be always the same. So we could talk about the eigenvalues, the eigenvalues of a linear transformation because for similar matrices, they have the same eigenvalues. We could talk about the determinant of a linear transformation. We could talk about the trace. We could talk about the rank of a linear transformation. These are words that often describe properties of a matrix, but we can also talk about these terms with respect to a linear transformation because it doesn't matter which matrix representation you choose, the eigenvalues don't change, the determinant doesn't change, the trace doesn't change. These are all things that are constant over the similarity class. And so as a final example right here, let's consider a map from R3 to R3, given by the following rule X, Y, and Z. If they're given, we can compute a vector in R3 by combining the vectors. You can see them right there. If you were to consider the matrix representation, this is the standard matrix representation for R3, then T looks like the matrix five, two, zero, three or two, three, negative one, second row and last row, 26, 16 and negative two. Sounds like a perfectly good matrix representation, right? Well, what if we come along and wanna use a non-standard basis? What if we wanna use this basis right here, the basis B, where the first vector is two, negative three, one, the second vector is negative one, two, two, and the third vector is negative one, one, negative two. What would be the matrix representation with respect to this non-standard basis? Well, like we'd have to compute before, like we had to do earlier, we have to figure out, what is, if we take AB1 and put into B coordinates, what does that look like? AB2, AB3, put them in B coordinates. So we have to row reduce the matrix right here, B augment the image of B when we hit it by matrix multiplication of A. So you see over here, this is our basis B. This is right here, this is the image of that basis B if you multiply it by A. And so again, I'm not gonna go through the details of row reduction, but if you row reduce it, well, B, since it's a basis, we'll row reduce to the identity. And then look over here, this reduces to a diagonal matrix whose diagonal entries are one, are two, one, and three right there. And so this diagonal matrix represents the matrix representation of T with respect to B coordinates. And you'll notice that this is a diagonal matrix and multiplying by a diagonal matrix is much, much easier than multiplying by that matrix we saw before. And why does this matrix representation provide something much simpler than the previous standard representation? Well, the standard representation came from using the standard basis E1, E2, E3. This matrix representation came from using an eigen basis. You'll notice that these vectors right here, vectors B1, B2, B3, this is an eigen basis. If you take A times B1, you end up with two times B1. If you take A times B2, you end up with one times B2. And if you take B3 and times it by A, you end up with three times B3. And so because you have a, since you have a eigen basis, what you're doing is instead of representing the matrix by its standard representation, you can instead represent the matrix using its diagonalization, this diagonal matrix. And from a computational point of view, working, since we can always change from one coordinate system to another, we can solve the problem with this easier basis system. To use another calculus analogy, in calculus two, sometimes you're approached with problems, in calculus three as well, sometimes you're approached with problems in one coordinate system, but it's like, hey, this problem's a lot easier to do in a different coordinate system. So like instead of using Cartesian coordinates, sometimes you switch over to polar coordinates or maybe you wanna use spherical coordinates or things similar to that, right? A cylindrical coordinates. You can change your problem to a more convenient coordinate system because instead of just a three by three matrix, which needs nine bits of information, we might actually be able to get away with just three bits of information in order to do the same work of that same transformation. This is why in calculus, we change variables all the time. And this is also why in calculus, especially multivariable calculus, when you change variables, the Jacobian comes into play. The Jacobian is the determinant of the change of basis matrix of the determinant, not the determinant of the derivative, excuse me, as you're switching from one coordinate system to another. So the change of basis matrix of the derivative is present, you see that. Why the determinant? Well, as you remember before, determinants measure area or volume or hypervolume in the various affine geometries you're in. And so as integrals are trying to calculate volumes, areas, et cetera, the determinant is the linear way of measuring area, volume, et cetera. And the integral is just the non-linear way of calculating these same problems you see in linear algebra. Because after all, calculus is the following thing. You take all the principles we've learned recently in linear algebra. You take the linear algebra things like linear transformations, vectors, matrices, all of that stuff. You take linear algebra and you add to it limits. And this is what you mean by calculus. And so this is sort of like the secret, the number 23 so to speak, when it comes to calculus, what's the meaning of life? Well, calculus is just you take linear problems and you add limits to it. And therefore you can solve various non-linear problems with the technique of calculus. And so I wanna end our course with this sort of connection to calculus because it's a really nice thing to see. I mean, linear algebra is great because of the many algebraic and geometric applications we've seen here. But to see like the bigger picture, many people learn these linear algebraic tools because they're gonna continue deeper and deeper into calculus. And you really cannot divorce calculus from linear algebra in the same way that you couldn't divorce calculus from limits. You need the two processes together to really master the types of problems we get in numerical analysis, differential equations, statistics, et cetera. And so that brings us to the end of this video and to the end of our course. Great at you for everyone who made it all the way to the end right here. It was good to have you. As always, if you have any questions, feel free to post them in the comments below on this video or any of the videos you've been watching. Please like this video, subscribe if you wanna see cool math videos in the future. Just because you've come to the end of your first semester of linear algebra doesn't mean your mathematical journey has ended. Oh please, just let it begin. Let's meet up in some future videos and learn some other cool Catan tech next time. I hope to see you then. Have a great day everyone, bye.


Hi, this is Corey Morris from Alhous University and we're here today at Mobile Think in Alhous, Denmark. Today we have the pleasure of speaking with Lars Vestergaard. Lars is one of our alumni. Lars is going to tell us a little bit more about his company and what he does. Lars, thank you so very much for agreeing to be with us today. Why don't we just start with one of the basic questions. Tell us a little bit about yourself. My name is, as you mentioned, Lars Vestergaard, and that is a Danish name. So I actually went to ASB and finished there in 1998. I graduated with a degree in international business, a Master of Science in international business. And right now I work for a Danish software company in the mobile business called Mobile Think, where I work as the managing director of a business unit. Can you tell us a little bit more about Mobile Think? For example, what do you guys do? Mobile Think is a mobile software company that works with a piece of software that is necessary for mobile operators around the world to control devices. It's a device management software that if you travel and buy a local SIM card and insert into your mobile phone, allows you to do data. So if you go anywhere and want to do something beyond voice and SMS, you need our solution to get it going. No end user ever knows about it or sees it.  So you're the guys who do that? We do that. We ensure that your holiday is cheap and that you can get the information you want on your mobile. We like that. So Lars, in three words, how would you describe Mobile Think? Mobile Think is a very young, it's a very enthusiastic culture. And the three words that we would use is trust, freedom and then lack of resources. The first two relate to part of our mission because really employees who work here feel that they can be trusted, feel the freedom because you can come and go as you want and you really are in a situation which is well adjusted, I guess. It's well suited to the audience of new grads. When it comes to the lack of resources it's because we're a small company, we're 60 employees and we could always do better in terms of our software. So you're the managing director. What does a managing director at Mobile Think actually do? I was brought on as a managing director to build up a proposition that was almost in the market and yet it wasn't. So I was here to revise the product internally. I was here to do all the necessary bits when it came to the PR, to the marketing, to the building of a website and taking the proposition out to market. So really a selling managing director role. And how did you get the job? That's a great networking story but I know the chairman and have known him for ten years and I met him at a large industry trade fair a few years ago and six months later we were discussing a contract. Now Lars, I know you've made lots of good decisions to get to where you are today but what is probably the best decision you've ever made business wise? That was when I in 2001 was offered a job in London and made that move. That was really the best decision that I ever made. Let's flip that and talk about some of the most difficult decisions or maybe the most difficult decision you've ever had to make in business. I think the most difficult decision relates to the best decision so that was actually leaving London was really a hard decision. The other one, if anything, was to admit defeat. So I was working out in Singapore in 1998 and the currency crisis had killed the company that I was working for and I had really, really set my nose up to a career in Asia Pacific for a number of years and seeing the end to that that I wouldn't want to take a local position was quite the defeat and returning back home in shame back then was really tough. Let's go back and look at your role as managing director. In your opinion, what's the best way to motivate your team? I have dealt with many teams in the past where you can let it depend on who that person is. So some people are very driven by reward and that can be a tap on the shoulder. It can be an appraisal that says you've done a good job, we've made a difference and for others it's pure monetary. So it's about having the right paycheck, having the right annual increase in your salary and generally, even though in Denmark we do focus a lot on how we have a nice time here and we leave at four o'clock and forget about bonuses, it's not really part of it. Money does rock this world, money does matter. Okay, let's jump into a time capsule, dial back the dates and look back at your time at ASB. And I'd like for you to talk about how ASB prepared you for this jump into the business world. If anything really made a difference for me it was the fact that I was allowed to go in and compliment a very theoretical ASB education at that time with something more practical. So I actually chose to go and study abroad. I went on one of the relationships that ASB at the time had with the university in Melbourne, RMIT, Royal Melbourne Institute of Technology and I took a semester there that was very practically founded. So we had for example one course where a business leader a week came in and gave us an idea about his or her world and we had a chance to discuss that. So what was ASB like back in the 90s? I won't lie to you and say that it was an undisputed pleasure because it was hard work. It really needed a lot of focus. From seeing a large group of people in year one to realizing that at the end of it half would be sorted away was actually quite tough. I wouldn't say that you only survived if you used your elbows but you really had to apply a certain amount of emphasis on. It's not all about drinking, it's not all about having a great social life. This was serious. If you really wanted this you needed to read your books and do well in the exams. So from that point of view it was tough work. Can you talk a little bit more about what it was like actually living and working abroad? How has that been for you? The studying abroad started quite early because I went to high school when I was a teenager but those two semesters I ended up having because I took one semester of my Kent Mac my master degree in Australia and I actually wrote my dissertation in Singapore Institute of Management. There was a great start to actually see what international life was like. I've managed to live and work in several countries and the nine years that I was working out of London was an undisputed positive thing. The fact that you were at the centre of excellence among a lot of peers in your industry has really shaped where I am today. Lars, what does it mean to you to be an alum from the university? To be a form of student from the university. Being an alum at the university is a new concept that only came to mind or to play for me a few years ago. Right now I use it a lot. I didn't miss it when I left the university. Back in those days it wasn't a big issue. I guess it was a big American issue at the time but it wasn't something that I felt I was missing. But having seen what it has been developed to be and what you can do with it, how you can associate yourself with the university, you can with pride mention where you have studied and where you come from is actually a very important thing. And I have certainly used it a lot. Speaking of students, is there any advice that you'd like to offer current students or even prospective students? I think it's a matter of really having that focus. Allow part of you to get driven away with what it is you want to do but allow part of you to also stay open to suggestions, stay open to chance, stay open to coincidence because that is what life really brings you. You can plan and apply focus but keep an open mind on what is going on. But in the first instance, don't party too much. Don't drink all the beers that all the rest do necessarily on the Friday and the Thursday because studying, who are we kidding? It's a task, it's a job. You have one life. Don't let business people tell you that you have a family time and you have a family life and then you have a business life and you get yourself a blackberry mobile phone so you can manage to go between the two. No, you have one life. So if you don't like the business life, leave it. You can make the changes and find something that really works for you that you can associate yourself with that allows you to be you in a family context and allows you to be you in a business context. Lars, thanks so very much for all the time and for talking with us. We really do appreciate it. And best of luck to you. Thank you very much. Once again, I'm Corey Morris with Ohos University. This has been Lars Vestegor with Mobile Think and thanks for your time.


Urban development must consider the interdependence of its various infrastructures in order to create sustainable and resilient systems. Transportation, water, and solid waste are highly interconnected, requiring an integrated approach to planning and design. Further research is needed to quantify these interdependencies and develop more effective strategies for managing them. This article was authored by Purnima A. J. Singha, Civil Durable and Lina Katan.