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}


$\gamma$ 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!

  • $\begingroup$ 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

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?


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.