2
$\begingroup$

We have a linear transformation $A: R³ \rightarrow R³$ where in a basis

$$ B = \{ \begin{bmatrix} 1\\ 2\\ 1 \end{bmatrix} \begin{bmatrix} 2\\ 1\\ 1 \end{bmatrix} \begin{bmatrix} 1\\ 0\\ 0 \end{bmatrix} \} $$

There is a matrix $$A_B=\begin{bmatrix} 2 & 0 & 0 \\ 0 & 1 & 0\\ 0& 0 & -1 \end{bmatrix}$$

We need to find this matrix in the standard basis ( this means vectors that have only one $1$ element and all others 0, so that they are independent)

I tried the following:

I would multiply:

$$A_B*\begin{bmatrix} 1\\ 0\\ 0 \end{bmatrix}=\begin{bmatrix} 2\\ 0\\ 0 \end{bmatrix}$$ Then I would try to get this matrix from $x*e_1+y*e_2+z*e_3$.

Where I would put $x,y$ and$ z$ into it's own vector which should be the first column in the new matrix that is based on the standard basis.

The thing is, I get the same matrix ($A_B$). I know that this is how we would calculate if we would have the matrix already in a standard basis and would want to write it in another matrix. Why doesn't this work the other way around. Did I miss something? How to solve this problem then?

$\endgroup$
1
1
$\begingroup$

Note that $$A_{BC} = \begin{bmatrix} 1 & 2 & 1 \\ 2 & 1 & 0 \\ 1 & 1 & 0 \end{bmatrix}$$ Is the change of basis matrix from $C$ to $B$ where $C=\{(1,0,0),\;(0,1,0),\;(0,0,1)\}$ and $B=\{(1,2,1),\;(2,1,1),\;(1,0,0)\}$. Then the matrix you are looking for is $A_{BC}\times A_B \times (A_{BC})^{-1} $

Please, take a look to this question.

$\endgroup$

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.