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I am working on the question below, and would like some help checking my approach to the problem.

Suppose that $T: \mathbb{R}^3 \to \mathbb{R}^2$ is given (in the usual coordinates) by the matrix $$A= \begin{bmatrix} 2 & 1 & 5\\ 1 & 1 & 3 \end{bmatrix}$$ If $\beta = \left \langle (1,1,1),(2,0,1),(3,2,1) \right \rangle$ is a new basis in $\mathbb{R}^3$, and $\alpha = \left \langle (3,5),(1,2) \right \rangle$ a new basis in $\mathbb{R}^2$, find the matrix for $T$ with respect to the new basis on both sides.

So, I am given $A\mathbf{v}=\mathbf{v'}$, where $\mathbf{v}$ is the input in the standard basis, and $\mathbf{v'}$ is the output in the standard basis. I need to be able to input a vector in basis $\beta$, perform the transformation, and receive output in basis $\alpha$.

If $B$ is the matrix that takes $\beta$ to the standard basis (S), then $B^{-1}$ is the matrix that takes the standard basis to $\beta$. Similarly, if $Z$ is the matrix that takes $\alpha$ to the standard basis, then $Z^{-1}$ is the matrix that takes the standard basis to $\alpha$.

I believe the steps I need are as follows: (1) Convert an input from $\beta$ into the standard basis; (2) Perform the transformation; (3) Convert the output from the standard basis into $\alpha$.

So, let $\mathbf{x}$ be an input vector in $\beta$, then $B\mathbf{x}=\mathbf{v}$, and

$$AB\mathbf{x}=\mathbf{v'}$$

Now I have input from $\beta$, converted to $S$, and put through the transformation with the output given in the standard basis. I still need to convert the output into $\alpha$, so I write

$$Z^{-1}AB\mathbf{x}=Z^{-1} \mathbf{v'}$$

because the matrix $Z^{-1}$ changes the vector from $S$ to $\alpha$. So the matrix for $T$ with respect to the new basis on both sides would be given by $Z^{-1}AB$. That is,

$$ Z^{-1}AB= \begin{bmatrix} 2 & -1\\ -5 & 3 \end{bmatrix} \begin{bmatrix} 2 & 1 & 5\\ 1 & 1 & 3 \end{bmatrix} \begin{bmatrix} 1 & 2 & 3\\ 1 & 0 & 2\\ 1 & 1 & 1 \end{bmatrix} $$

Is my reasoning correct? Thanks for any input.

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Yes, basically correct. The only thing you exchanged is $B$ and $B^{-1}$, as you introduced $B$ in the last line, it consists of the elements of $\beta$ as column vectors, so $B$ corresponds to the linear transformation that takes the standards basis to $\beta$ (coordinated w.r.t the standard basis on both sides). Else $Z^{-1}AB$ is correct.

You can verify your result by calculating $Ab_1,Ab_2,Ab_3$ for the elements $b_i$ of $\beta$ (this is anyway just the matrix $AB$), and then expressing these 2d vectors by the elements of $\alpha$. For example $Ab_1=\pmatrix{8\\5}=11\cdot\pmatrix{3\\5}-25\cdot\pmatrix{1\\2}$, so the first column will be $\pmatrix{11\\-25}$...

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