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linear algebra is quite confusing at the beginng... So:
Let $V=lin \left ( (1,1,3,2),(4,5,2,5),(2,3,-4,1),(1,2,-5,5) \right ) \subset \mathbb{R}^4$
a) Find base and dimension of $V$
b)For what values of $r\in\mathbb{R} $ exists base $\alpha_1, \alpha_2,\alpha_3,\alpha_4 $ in $\mathbb{R}^4$ which satisfies: $\alpha_1, \alpha_2 \in V$ and vector $\beta=(1,1,3,3)$ has in that base coordinates $0,1,r,0$? For every such $r$ give example of base.

First part of quest is easy, just enough to pout everything in matrix and calculate.
Second part is not easy. I don't really understand it. I calculated that $\dim V=2$, so I have three vectors to choose from, but I have to choose only two, so can I choose two most suitable? Further, when I choose two and add $\beta$ I still have only three, but i need for. Can I add one more unique vector to those three (as I am adding to $\mathbb{R}^4)$, e.g. $(0,0,0,1)$?
And what with $r$? It really confuses me, any help and hints will be appreciated, thanks in advance.

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Matlab tells me that $\mathbb{dim} V = 3$. –  a.r. Nov 21 '12 at 16:52

2 Answers 2

up vote 1 down vote accepted

a) you just need to see is if the vector given are independent. they are not as $(4,5,2,5)$ is an addition of $2 \cdot (1,1,3,2)$ and $(2,3,-4,1)$ then check the rest are independent by computing the rank of the matrix:

1 1 3 2

2 3 -4 1

1 2 -5 5

thus you get a degree of 3, and those vectors are also Basis.

Now for 2, you simply got $dim(V)$ wrong, it is 3 (as explained above). try now

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But i have to choose two vectors from V, does it matter which? –  fdhd Nov 21 '12 at 17:07
Because you need to know what you are choosing from. For example after fixing the first section you can always just take $\alpha_2=(1,1,3,2) , \alpha_1=(1,2,-5,5),\alpha_3=(0,0,0,1)$ and $\alpha_4=(0,0,1,0)$ and then you have the base coordinates 0 1 1 0. –  user44874 Nov 21 '12 at 17:13
Alright, i think i got it, thanks. One more question, is a base of space somehow related to the base of subspace of system of solutions? In particular: are they equal? –  fdhd Nov 21 '12 at 17:23
what do you mean by equal? the two basis $(2,-3) (1,1)$ and $(1,0),(0,1)$ are both basis of the same space $\mathbb{R}^2$ but they are not "equal" they simply span (produce) the same space. Also, what system of solutions are you referring to? can give an example? –  user44874 Nov 21 '12 at 17:34
umm.. system of solutions of, e.g. 3 equations $x_1,x_2,x_3,x_4$, by equal i mean that the spaces are equivalent. I ask if base of that equations span the same space as base of subspace of solutions of this equations –  fdhd Nov 21 '12 at 17:37

b) Let's say you are given a real number $ r $. Now, you have to check if it possible to find 4 linearly independent vectors such that at least two of them are in the space V. No condition on the remaining two vectors except linear independence. You take these 4 vectors $ a_1, a_2, a_3, a_4 $ and try to express the vector $ \beta $ in terms of them. It means, that you find 4 numbers $ c_1, c_2, c_3, c_4 $ such that $ \beta = c_1a_1 + c_2a_2 + c_3a_3 + c_4a_4$. If, as in the question, the numbers are 0, 1, r and 0, these $ a_i $ are "ok".

Finally, what you are supposed to do in the exercise - you have to find for which numbers $ r $ it is possible to find these vectors $ a_1, a_2, a_3, a_4 $ so that they are "ok"(when you express $ \beta $ in terms of them, the coefficients are 0, 1, r, 0).

Hope that helps :)

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