# Linear transformation from a standard basis to another basis

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?

## 1 Answer

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.