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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 $\mathbb{dim} V = 3$.2012-11-21

2 Answers 2

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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?2012-11-21
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    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 0.2012-11-21
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    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?2012-11-21
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    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?2012-11-21
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    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 equations2012-11-21
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    First, you do not say "equal base". What you mean is "equal space" you are asking if the space of the solutions is equal to the space of the equations. not sure what "space of the equations" mean but i will try to answer in the following comment2012-11-21
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    Again, i am having difficulty understanding exactly what you mean. I never heard of "base of equation span"; Lets take a specific example: if we take the matrix A to 2x2 with the first raw 1 0 and the second 0 0. Then the solution space for the system of equations Ax=0 is {(0,1)}. "The base of equations" however is {(1,0)} (again not sure if i understood you correctly).2012-11-21
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    By base of subspace of solutions (its directly translated form Polish) i mean something like $(x_1,x_2,x_3,x_3)=(3x_3 -x_4, x_3+2x_4,x_3,x_4)=x_3(3,1,1,0) + x_4 (-1, 2,0,1)$ (just an example) So "base of solutions of..." here is $(3,1,1,0), (-1,2,0,1)$. Is it complementary with the base of main space, i mean base, which i get directly from matrix?2012-11-21
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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 :)