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This is the text I've been given for the test:

Let L be a linear map $\mathbb {R}^2 \to \mathbb{R}^2$, whose matrix respect to the standard basis is:

[L] = \begin{bmatrix} 1 & -1 \\[0.3em] -1 & 0 \\[0.3em] \end{bmatrix}

Now, consider the following two bases of $\mathbb R^2$:

$\beta$ V1 = \begin{bmatrix} 0 \\[0.3em] 1 \\[0.3em] \end{bmatrix}

$\beta$ v2 = \begin{bmatrix} , 1 \\[0.3em] 1 \\[0.3em] \end{bmatrix}

and

$\gamma$ w1 = \begin{bmatrix} , 1 \\[0.3em] 2 \\[0.3em] \end{bmatrix}

$\gamma$ w2 = \begin{bmatrix} , 0 \\[0.3em] 1 \\[0.3em] \end{bmatrix}

Command: write the associated matrix L with respect at base $\beta$ (domain) to $\gamma$ (codomain).

Solution: \begin{bmatrix} -1 & 0 \\[0.3em] 2 & -1 \\[0.3em] \end{bmatrix}

I thought i could write (1,0) as the difference between v1 and v2, but then i don't when or how to use it. I'm pretty confused on how i should relate the first matrix with the two basis!

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  • $\begingroup$ What is your definition of the matrix of $L$ wrt the bases $\beta$ and $\gamma$? Surely it tells you that you need to work out $Tb_1$ and $Tb_2$ and express these in terms of $c_1$ and $c_2$ and then you've got the answer ....? $\endgroup$ – ancientmathematician Apr 26 '17 at 9:53
  • $\begingroup$ That's what i'm not working out! I updated the text; i don't know how to use the first matrix with the two basis! $\endgroup$ – InfoB Apr 26 '17 at 11:12
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I can't use your notation.

Let $\beta=\{b_1, b_2\}$ be the two vectors you specify and $\gamma=\{c_1,c_2\}$ be the other two.

Then $$ Lb_1= \begin{pmatrix} 1 & -1\\ -1 & 0 \end{pmatrix} \begin{pmatrix} 0 \\ 1 \end{pmatrix}= \begin{pmatrix} -1\\ 0 \end{pmatrix}=-c_1+2c_2. $$

Hence $$ {}_\beta[L]_\gamma= \begin{pmatrix} -1 & \cdot \\ 2 & \cdot \end{pmatrix}. $$

So you can do the next column?

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