Linear Transformation and Basis Question I am working on a question and was looking for some help. The question is

Suppose that $\mathbf{v}_1 = (1,2)$, $\mathbf{v}_2 = (2,-1)$ and that the basis $\beta$ is $\beta = \left \langle \mathbf{v}_1 , \mathbf{v}_2 \right \rangle$. Let $T$ be the linear transformation from $\mathbb{R}^2$ to $\mathbb{R}^2$ given by $T(\mathbf{v}_1)=\mathbf{v}_1$ and $T(\mathbf{v}_2)= \mathbf{0}$.
(a) Write down the matrix for $T$ in the new basis $\beta$. (You should be able to do this directly from the definition of $T$.)
(b) Use this to write down the matrix for $T$ in the standard basis.

So, the matrix to convert a vector from $\beta$ to the standard basis is 
$$
\begin{bmatrix}
1 & 2\\ 
2 & -1
\end{bmatrix}
.$$
As such, the matrix to convert a vector from the standard basis to $\beta$ is 
$$
\begin{bmatrix}
1 & 2\\ 
2 & -1
\end{bmatrix}^{-1}
=
\begin{bmatrix}
1/5 & 2/5 \\ 
2/5 & -1/5
\end{bmatrix}
$$
Converting $\mathbf{v}_1$ and $\mathbf{v}_2$ into $\beta$ is done in the following manner,
$$
\begin{bmatrix}
1/5 & 2/5 \\ 
2/5 & -1/5
\end{bmatrix}
\begin{bmatrix}
1\\ 
2
\end{bmatrix}
=
\begin{bmatrix}
1\\ 
0
\end{bmatrix}
$$
$$
\begin{bmatrix}
1/5 & 2/5 \\ 
2/5 & -1/5
\end{bmatrix}
\begin{bmatrix}
2\\ 
-1
\end{bmatrix}
=
\begin{bmatrix}
0\\ 
1
\end{bmatrix}.
$$
Which, in the transformation, gives $T_{\beta} (1,0) = (1,0)$ and $T_{\beta}(0,1)=(0,0)$. So the matrix for $T$ in the basis $\beta$ is
$$
A=
\begin{bmatrix}
1 & 0 \\ 
0 & 0
\end{bmatrix}.
$$
Now, I am not sure if what I have done is correct. In any case, how do I proceed with (b)? Thanks for any help. I think I am a bit mixed up.
 A: The answer you've gotten for (a) is correct, but you did a lot of unnecessary work.  If $T(v_1) = v_1$ then you know the first column of $A$ contains the coefficients in the basis expansion of $v_1$.  Since $v_1$ is in your basis this expansion is just $v_1$ and so the first column is $\begin{bmatrix} 1 \\ 0\end{bmatrix}$.  Similarly the second column must be zero.
Now once you've gotten $A$ you change basis, so just compute $BAB^{-1}$ where $B$ is the change of basis matrix you've already found.
A: What you have right now is this:
$$(\text{$A$ in $\beta$ basis})(\text{input vector in $\beta$ basis}) = \text{output vector in $\beta$ basis}$$
Let me just write this as
$$A v= v'$$
where $v, v', A$ are all understood to be in the $\beta$ basis.
You have the matrix $B$ that converts from $\beta$ basis to standard basis.  Multplying by that on the left will convert the output vector to standard basis.  But you also want to input a vector in standard basis.  You can accomplish this by inserting a $BB^{-1}$ between $A$ and $v$ like so:
$$B(AB^{-1}[Bv]) = Bv'$$
$Bv$ is an input vector in the standard basis.  $Bv'$ is an output vector in the standard basis.  Therefore, $BAB^{-1}$ converts $A$ from $\beta$ basis to standard basis.
