Describing the action of T (linear transformation) on a general matrix I am not familiar with linear transformations in general, and as such, I do not know how to approach this type of question as the examples I'm given/looked up online usually deal with finding the transformation matrix itself. 
Suppose $T:M_{2,2}\rightarrow P_{3}$ is a linear transformation whose action on the standard basis for $M_{2,2}$ is as follows:
$$T\begin{bmatrix}
1 & 0\\ 0
 & 0
\end{bmatrix}= x^3-3x^2+x-2$$
$$T\begin{bmatrix}
0 & 1\\ 0 & 0\end{bmatrix}= x^3-3x^2+2x-2$$
$$T\begin{bmatrix}
0 & 0\\ 1
 & 0
\end{bmatrix}=x^3-x^2+2x$$
$$T\begin{bmatrix}
0 & 0\\ 0
 & 1
\end{bmatrix}=3x^3-5x^2-1$$
I am asked to describe the action of $T$ on a general matrix using $x$ as a variable for the polynomial and $a,b,c,d$ as constants. I am assuming that I need to form some sort of expression in polynomial form. 
Any help would be greatly appreciated!
 A: $\begin{bmatrix}
a & b\\ 
 c& d
\end{bmatrix}=a\begin{bmatrix}
1 & 0\\ 0
 & 0
\end{bmatrix}+b\begin{bmatrix}
0 & 1\\ 0
 & 0
\end{bmatrix}+c\begin{bmatrix}
0 & 0\\ 1
 & 0
\end{bmatrix}+d\begin{bmatrix}
0 & 0\\ 0
 & 1
\end{bmatrix}$
Now since $T$ is linear so 
$T\begin{bmatrix}
a & b\\ 
 c& d
\end{bmatrix}=aT\begin{bmatrix}
1 & 0\\ 0
 & 0
\end{bmatrix}+bT\begin{bmatrix}
0 & 1\\ 0
 & 0
\end{bmatrix}+cT\begin{bmatrix}
0 & 0\\ 1
 & 0
\end{bmatrix}+dT\begin{bmatrix}
0 & 0\\ 0
 & 1
\end{bmatrix}$
now I hope you can take it from here!
A: As it turns out, I have been overthinking this problem.
 Taking $T=\begin{bmatrix}
1 & 1 & 1 & 3\\ 
-3 & -3 & -1 & -5\\ 
1 & 2 & 2 & 0\\ 
-2 & -2 & 0 & -1
\end{bmatrix}$ and multiplying it with the vector of constants $\begin{bmatrix} a\\b\\c\\d\end{bmatrix}$ gives me $$\begin{bmatrix} a+b+c+3d\\-3a-3b-c-5d\\a+2b+2c\\-2a-2b-d\end{bmatrix}$$
From here, row 1 corresponds with $x^3$, row 2 with $x^2$, row 3 with $x$, and row 4 with a constant of 1 (given the standard basis for the polynomials to be $1, x, x^2, x^3$, this makes sense).
Therefore, the action of $T$ on the general matrix can be written in polynomial form as $$T\begin{bmatrix} a&b\\c&d\\ \end{bmatrix}=(a+b+c+3d)x^3+(-3a-3b-c-5d)x^2+(a+2b+2c)x+(-2a-2b-d)$$
Alternatively, another method (suggested to me previously) is to simply take $$aT\begin{bmatrix} 1&0\\0&0 \end{bmatrix}+bT\begin{bmatrix} 0&1\\0&0 \end{bmatrix}+cT\begin{bmatrix} 0&0\\1&0 \end{bmatrix}+dT\begin{bmatrix} 0&0\\0&1 \end{bmatrix}$$ such that you get
$$a(x^3-3x^2+x-2)+b(x^3-3x^2+2x-2)+c(x^3-x^2+2)+d(3x^3-5x^2-1)$$
Combining like terms, we end up with the same result as above. Both methods are essentially the same.
