0
$\begingroup$

The question is as follows.

Let $\alpha = \left\{ \begin{bmatrix}1&0\\0&0 \end{bmatrix},\begin{bmatrix}0&1\\0&0 \end{bmatrix},\begin{bmatrix}0&0\\1&0 \end{bmatrix},\begin{bmatrix}0&0\\0&1 \end{bmatrix} \right\}$.

If $A =\begin{bmatrix} 1 & -2 \\ 0 & 4 \end{bmatrix} $, compute $[A]_\alpha$.

How do I do this? The other problems in this section define a linear transformation and ask you to find the matrix with respect to a given basis. Here is a matrix and we must find another matrix? The answer is $(1,-2,0,4)^T$, but I'm not sure why exactly this is the case.

$\endgroup$
  • 2
    $\begingroup$ $[A]_\alpha$ is the coordinate vector of $A$ with respect to the basis $\alpha$. So how can you make $A$ as a linear combination of the basis vectors? You take $1$ of the first one, $-2$ of the second, $0$ of the third, and $4$ of the fourth, which translates to the coordinate vector $(1,-2,0,4)$. $\endgroup$ – Dave Jun 8 '17 at 1:43
  • 2
    $\begingroup$ If your 4 matrices are named $M_1,M_2,M_3,M_4$, it means $A=(1)M_1+(-2)M_2+(0)M_3+(4)M_4$, that's all. $\endgroup$ – Jean Marie Jun 8 '17 at 1:43
3
$\begingroup$

This is intuitive. They are clearly looking for a linear combination of your basis matrices that equal the matrix $A$. Clearly this is

$$A =(1)\begin{bmatrix}1&0\\0&0 \end{bmatrix} -2 \begin{bmatrix}0&1\\0&0 \end{bmatrix} + (0)\begin{bmatrix}0&0\\1&0 \end{bmatrix} + (4)\begin{bmatrix}0&0\\0&1 \end{bmatrix} .$$

$\endgroup$

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.