Linear Algebra: Basis Question 
Let $b_1 = \left(\begin{matrix}-1 \\ -3\end{matrix}\right)$ and $b_2 = \left(\begin{matrix} -3 \\ -10\end{matrix}\right)$. The set $B = \{b_1, b_2\}$ is a basis for $\mathbb{R}^2$. Let $T: \mathbb{R}^2 \to \mathbb{R}^2$ is a linear transformation such that $T(b_1) = 4b_1 + 4b_2$ and $T(b_2) = 4b_1 + 5b_2$. Then the matrix of $T$ relative to the basis $B$ is
$$[T]_B = \left(\begin{matrix} \color{red}{4} && \color{red}{4} \\ \color{red}{4} && \color{red}{5} \end{matrix}\right),$$
and the matrix of $T$ relative to the standard basis $E$ for $\mathbb{R}^2$ is
$$[T]_E = \left(\begin{matrix} \color{red}{?} && \color{red}{?} \\ \color{red}{?} && \color{red}{?}\end{matrix}\right).$$

I managed to figure out the first part. I think that the standard basis $E$ would be $e_1 = (1,0)$ and $e_2 = (0,1)$ but I don't know how to use this to solve the second half of the question.
 A: Recall what the matrix $[\, T\,]_B$ is: 
If you write a vector $x\in\Bbb R^2$ in terms of the basis $B=\{\,b_1,b_2\,\}$
$$
x=\alpha_1 b_1+\alpha_2 b_2,
$$
then if you multiply $[\, T\,]_B$ by the coordinate vector $x_B=\bigl[{\alpha_1\atop\alpha_2}\bigr]$, you get the coordinate vector of $T(x)$ with respect to $B$.
That is
$$\tag{1}
[T(x)]_B = [\,T\,]_B x_B.
$$

Now  the columns of the matrix $[\,T\,]_E$ where $E=\{e_1,e_2\}$ are the vectors $T(e_1)$ and $T(e_2)$.  To find these vectors, we can use the matrix  $[\, T\,]_B$.
There are three steps involved here. Considering the vector $e_2$, we have to


*

*Find the coordinates of $e_2$ with respect to the basis $B$.

*Find the coordinates of $T(e_2)$ with respect to the basis $B$.

*Find $T(e_2)$ expressed in the standard basis.


Step 1: For $e_2=(0,1)$, we first find the coordinates of $e_2$ in terms of the basis $B$. Towards this end, we have to solve the system
$$
\Bigl[{0\atop1}\Bigr]= \alpha_1 \Bigl[{-1\atop-3}\Bigr] +\alpha_2\Bigl[{-3\atop-10}\Bigr].
$$ 
Doing so gives:
$$\alpha_1=3,\quad \alpha_2=-1$$ The coordinate vector of $e_2$ with respect to $B$ is 
$\bigl[{3\atop-1}\bigr]$. 
Note we could have done this differently: the coordinate vector  $\bigl[{\alpha_1\atop\alpha_2}\bigr]$ of $x$ with respect to $B$ satisfies $[b_1 \ b_2] \bigl[{\alpha_1\atop\alpha_2}\bigr] =x$; so $\bigl[{\alpha_1\atop\alpha_2}  \bigr]=[b_1 \ b_2]^{-1}  x$.
Thus we could have found $[b_1\ b_2]^{-1}$ and just multiplied this by $e_2$. This is actually preferable, since we can use the inverse when considering $e_1$ later.  

Step 2:
Using $(1)$ now, the coordinate vector of $T(e_2)$ with respect to $B$ is
$$
[\,T\,]_B (e_2)_B =
\Bigl[ \matrix{4&4\cr 4&5 }\Bigr]\Bigl[{3\atop -1} \Bigr] =
\Bigl[{8\atop 7} \Bigr].
$$

Step 3:
But note that $T(e_2)$ is not the vector $\bigl[{8\atop 7} \bigr]$; this vector gives the coordinates of $T(e_2)$ with respect to the basis $B$. In general, if $x_B$ is the coordinate vector  of $x$ with respect to $B$, then $x=[b_1\ b_2] x_B$;
so $$
T(e_2)=[b_1\ b_2]\Bigl[{8\atop 7} \Bigr] =8\, b_1+7\, b_2=  8\Bigl[{-1\atop -3} \Bigr] +7 \Bigl[{-3\atop -10} \Bigr]
=\Bigl[{-29\atop -94} \Bigr].
$$

Thus, the second column of $[\,T\,]_E$  is $\bigl[{-29\atop -94} \bigr]$.  
To find the first column
 of $[\,T\,]_E$, apply the same procedure to the vector $e_1$.  The first step here would be to write $e_1$ in terms of the basis $B$. To do that, you need to solve the system
$$
\Bigl[{1\atop0}\Bigr]= \alpha_1 \Bigl[{-1\atop-3}\Bigr] +\alpha_2\Bigl[{-3\atop-10}\Bigr].
$$ 
or compute:
$$\Bigl[{\alpha_1\atop\alpha_2}  \Bigr]=[b_1 \ b_2]^{-1}  e_1.$$



In general, given $x$ written in terms of the standard basis, the coordinates of $x$ with respect to $B$ are given by $ P^{-1} x$ where $P=[b_1\ b_2]$. Then the coordinates of $T(x)$ with respect to $B$ are $[\,T\,]_B P^{-1} x$.  Then we have that  $T(x)$ expressed in terms of the usual basis is $P [\,T\,]_B P^{-1} x$. So,
$$
[\,T\,]_E = P[\,T\,]_B P^{-1}.
$$
A: Recall theorem :We know that $[T]_E=P[T]_BP^{-1}$ where $P=[\operatorname{id}]_{E,B}=[b_1 \,\, b_2]$.
Now you can easily solve it.
A: Let's think about this.
Well we need the values of $T(e_1), T(e_2)$ in terms of $e_1,e_2$.
So first we are going to have to write $b_1, b_2$ in terms of $e_1,e_2$.
If we do this and use it in our definition of $T$ then we see that:
$ -T(e_1) - 3T(e_2) = -16e_1 -52e_2$
$ -3T(e_1) - 10T(e_2) = -19e_1 -62e_2$
We can now solve these equations for $T(e_1), T(e_2)$ and get:
$T(e_1) = 103e_1 +334e_2$
$T(e_2) = -29e_1 - 94e_2$
From this you get the matrix of $T$ with respect to the standard basis.
