Finding the Matrix of a Given Linear Transformation T with respect to a basis I've been scouring the web and my textbook to no avail. I understand how to transform a matrix with respect to the standard basis, but not to a basis comprised of matrices.
I need to find the matrix of T with respect to a basis. My transformation matrix is:
$$ 
A =
\begin{bmatrix}
1  &   2   \\
0  &   3   \\       
\end{bmatrix} $$
and I need to find it with respect to the basis:
$$
B = \left\{
\begin{bmatrix}
1& 0  \\
0  &0  \\       
\end{bmatrix} ,
\begin{bmatrix}
0 &  1 \\
0 &  0  \\       
\end{bmatrix},
\begin{bmatrix}
0 &  1  \\
0  & 1  \\       
\end{bmatrix}\right\}
$$
I'm sorry my formatting is rough, the three B matrices are in the book as B = (Matrix 1, Matrix 2, Matrix 3).
I really don't understand this. My book has a column by column method that has completely stumped me.
 A: I think your difficulty is purely "conceptual". A vector space is ANY set that obeys the axioms of a vector space (such a definition assumes an associated "field of scalars"). In particular, over a given field $F$, the set:
$\text{Hom}_F(U,V)$ = all linear transformations $U \to V$ is a vector space.
So "matrices" are vectors, too! Look, we can add them: if $A = (a_{ij}), B = (b_{ij})$ then $A + B = (c_{ij})$, where for each $i,j: c_{ij} = a_{ij} + b_{ij}$, and we can "multiply by a scalar":
$rA = (ra_{ij})$ (sometimes this is written as $(rI)A$).
One way to "ease the conceptual transition" is called the "vectorization" of matrices: we just string the columns "head-to-toe" into one long column, so that:
$A = \begin{bmatrix}1&2\\0&3\end{bmatrix}$ becomes:
$A = \begin{bmatrix}1\\0\\2\\3\end{bmatrix}$, transforming an element of $\text{Mat}_2(F)$ into an element of $F^4$ (you can take $F = \Bbb R$, for concreteness, if you primarily deal with real vector spaces).
Seen this way, it becomes clear that the second coordinate of your basis vectors is "unnecessary baggage", as it is always $0$. This is no different than identifying the subspace:
$U = \{(x,0,y,z) \in \Bbb R^4\}$ with $\Bbb R^3 = \{(x,y,z):x,y,z \in \Bbb R\}$
It should be clear that $\phi:U \to \Bbb R^3$ given by $\phi(x,0,y,z) = (x,y,z)$ is a bijective linear transformation.
So, in your situation, you want to find $a,b,c$ such that:
$A = \begin{bmatrix}1\\0\\2\\3\end{bmatrix} = a\begin{bmatrix}1\\0\\0\\0\end{bmatrix} + b\begin{bmatrix}0\\0\\1\\0\end{bmatrix} + c\begin{bmatrix}0\\0\\1\\1\end{bmatrix}$
Which is just a system of 3 linear equations in 3 unknowns:
$a = 1\\b+c = 2\\c = 3$
that you should be proficient in solving by now.
A: From the nature of your question, I'm not sure you've transcribed it correctly for us. I suspect that the answer is this:
$$
\begin{bmatrix}
{1}  & {2}    \\
{0}  & {3}    \\       
\end{bmatrix}
= 
1 \cdot \begin{bmatrix}
{1}  & {0}    \\
{0}  & {0}    \\       
\end{bmatrix} + 
(-1) \cdot \begin{bmatrix}
{0}  & {1}    \\
{0}  & {0}    \\       
\end{bmatrix} + 
3 \cdot \begin{bmatrix}
0  &1    \\
0  & 1   \\       
\end{bmatrix}
$$
so that the coordinates of your matrix, $A$, with respect to the basis given, are $1, -1, 3$. Perhaps you're supposed to write that as a vector, in which case it'd be 
$$
\begin{bmatrix}
1    \\
-1   \\       
3
\end{bmatrix}.
$$
