A strange identity matrix? In class we were discussing plugging linear operators into polynomials to do different things, and we used $2x^2 + x - 4$ for the experiment, and plugged in $T$ where $T(\begin{bmatrix}
a & b\\ 
c & d
\end{bmatrix})=\begin{bmatrix}
0 & a\\ 
0 & d
\end{bmatrix}.$ Plugging in was generally straightforward, but when it got to the $4,$ it was changed to $4I = \begin{bmatrix}
4a & 4b\\ 
4c & 4d
\end{bmatrix}.$ Could you explain why this matrix is the identity matrix given the polynomial and the linear operator? Many thanks.
 A: Your operator $T$ acts on $2 \times 2$ matrices, so the number 4 should be replaced by $4I$ where $I$ is the identity operator on that space, namely the operator that doesn't change the matrix at all:
$$I(\begin{bmatrix}
a & b\\ 
c & d
\end{bmatrix})=\begin{bmatrix}
a & b\\ 
c & d
\end{bmatrix}.$$
(Since the space $V$ of $2\times 2$ matrices is four-dimensional, the matrix of this operator with respect to some basis for $V$ will be the usual $4 \times 4$ identity matrix diag(1,1,1,1).)
EDIT:
You're looking at linear operators on the vector space $V$ consisting of $2 \times 2$ matrices. So the "vectors" in your space are actually matrices, but they can be written as column vectors if you introduce some basis. For example, take
$$
e_1 = \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix},
\quad
e_2 = \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix},
\quad
e_3 = \begin{bmatrix} 0 & 0 \\ 1 & 0 \end{bmatrix},
\quad
e_4 = \begin{bmatrix} 0 & 0 \\ 0 & 1 \end{bmatrix}.
$$
Then
$$
\begin{bmatrix} a & b \\ c & d \end{bmatrix}
= a e_1 + b e_2 + c e_3 + d e_4,
$$
so that the coordinate vector of $\begin{bmatrix} a & b \\ c & d \end{bmatrix}$ (with respect to our chosen basis) is
$\begin{bmatrix} a \\ b \\ c \\ d \end{bmatrix}$.
Now the identity operator does nothing to the "vector"=matrix in $V$,
and this fact is expressed in terms of coordinate column vectors like this:
$$
\begin{bmatrix} 1&0&0&0 \\ 0&1&0&0 \\ 0&0&1&0 \\ 0&0&0&1 \end{bmatrix}
\begin{bmatrix} a \\ b \\ c \\ d \end{bmatrix}
=\begin{bmatrix} a \\ b \\ c \\ d \end{bmatrix}.
$$
So the matrix representing the identity operator in this basis (and in fact in any basis) is the $4 \times 4$ identity matrix.
For comparison, the action of the operator $T$ is represented in our basis as
$$
\begin{bmatrix} 0&0&0&0 \\ 1&0&0&0 \\ 0&0&0&0 \\ 0&0&0&1 \end{bmatrix}
\begin{bmatrix} a \\ b \\ c \\ d \end{bmatrix}
=\begin{bmatrix} 0 \\ a \\ 0 \\ d \end{bmatrix}.
$$
If we had chosen some other basis, we would have gotten some other $4 \times 4$ matrix here; the operator itself is independent of the choice of basis, but the coordinate representation in terms of a $4 \times 4$ matrix usually depends on the choice of basis (it's just that the identity operator is always represented by the identity matrix, regardless of basis).
