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The subspace $S \subseteq \mathbb{R}^{4}$ is spanned by the vectors $v_1= \begin{pmatrix} 1\\ 2\\ 3\\ 0 \end{pmatrix}, v_2= \begin{pmatrix} -1\\ 5\\ 7\\ -1 \end{pmatrix}, v_3\begin{pmatrix} 5\\ -4\\ -5\\ 2 \end{pmatrix}$

The basis is $B= \left\{ \begin{pmatrix} 1\\ 2\\ 3\\ 0 \end{pmatrix}, \begin{pmatrix} 0\\ 7\\ 10\\ -1 \end{pmatrix} \right\}$

For each vector $v_1,v_2,v_3$ determine the coordinates relating to the basis.

I'm not sure if it will be correct that's why I need to ask:

For each vector, we have to do the following (I start with $v_1$):

$a\begin{pmatrix} 1\\ 2\\ 3\\ 0 \end{pmatrix}+b\begin{pmatrix} 0\\ 7\\ 10\\ -1 \end{pmatrix}= \begin{pmatrix} 1\\ 2\\ 3\\ 0 \end{pmatrix}$

This is easy, let $a=1$ and $b=0$.

But what's the coordinate for $v_1$ now? Is it $\begin{pmatrix} 1\\ 0 \end{pmatrix}$? And is it correctly done like that (apply the same for the other vectors..)?

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    $\begingroup$ Yes, it's correct. Sometimes, to emphasize that these are the coordinates with respect to a certain (not the standard) basis, the name of the basis is added as a subscript: $\begin{pmatrix} 1 \\ 0 \end{pmatrix}_B$. $\endgroup$
    – zipirovich
    Mar 14 '17 at 17:05
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Yes, you solution is correct.

But notice that since the subspace is spanned by only two vectors, there maybe vectors that does not fall into this sub-space - and for those vectors, you'll notice that you linear functions do not have a solution.

Example: $a\begin{pmatrix} 1\\ 2\\ 3\\ 0 \end{pmatrix}+b\begin{pmatrix} 0\\ 7\\ 10\\ -1 \end{pmatrix}= \begin{pmatrix} 1\\ 5\\ 3\\ 0 \end{pmatrix}$

For the equation above, you cannot find a pair of (a,b) that satisfies all four rows. Thus $(1, 5, 3, 0)^T$ is not in the subspace spanned by $(1,2,3,0)^T$ and $(0,7,10,-1)^T$

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(apply the same for the other vectors..)?

Correct! To determine the coordinates of $v_2,v_3$ with respect to the basis $B = \{b_1, b_2\}$ you will have to solve two systems of linear equations:

$$ a \cdot b_1 + b \cdot b_2 = v_2 $$ and $$ a \cdot b_1 + b \cdot b_2 = v_3 $$ for the coordinates $a$ and $b$.

P.S.: Note that $b_2 = v_1 + v_2$...

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That’s right. To find the coordinates of each vector $v_k$, you need to solve an equation of the form $v_k=aw_1+bw_2$ ($w_1$ and $w_2$ are the basis vectors).

Recall that one way to solve such an equation is by representing it as an augmented matrix and row-reducing. If fact, you can solve all three equations “in bulk” by packaging all of the vectors into a single matrix: $$\left(\begin{array}{rr|rr}1&0 & 1&-1&5 \\ 2&7 & 2&5&-4 \\ 3&10 & 3&7&-5 \\ 0&1 & 0&-1&2\end{array}\right) \to \left(\begin{array}{rr|rr}1&0 & 1&*&* \\ 0&1 & 0&*&* \\ 0&0 & 0&0&0 \\ 0&0 & 0&0&0\end{array}\right).$$ The coordinates of the vectors relative to the new basis are found in the first two rows of each column. Just as you’ve already found, the coordinate vector of $v_1$ relative to $B$ is $(1,0)^T$. I’ve left the others for you to compute. If any of the vectors lie outside of $S$ (which isn’t the case here per the problem statement), it will have a non-zero entry in some other row of the reduced matrix.

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