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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.

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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).

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Isn't the property of an identity matrix that when you multiply it by some other matrix in the space, that other matrix doesn't change? How does this work here, does it have to do with the operator? Could please expand a bit more (step by step)? Sorry, I am not getting/seeing it for some reason. :( –  linalqqq Sep 28 '11 at 7:32
    
@linalqqq: OK, I edited the answer and tried to explain a bit more in detail. –  Hans Lundmark Sep 28 '11 at 7:52
    
I wrote out $$T\begin{bmatrix} a & b \\ c & d \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix} \begin{bmatrix} a & b \\ c & d \end{bmatrix} \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix} + \begin{bmatrix} 0 & 0 \\ 0 & 1 \end{bmatrix} \begin{bmatrix} a & b \\ c & d \end{bmatrix} \begin{bmatrix} 0 & 0 \\ 0 & 1 \end{bmatrix} $$ and then tried to display $T^2$ and $T^3$ applied to the same general $2\times 2$ matrix, in all their splendor. That would be four terms, and then eight terms. I decided that might not be the most intuitive way to procede. –  Michael Hardy Sep 28 '11 at 11:44
    
Thank you! This makes sense. –  linalqqq Sep 28 '11 at 16:16

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