Characteristic equation of a linear map Question: Suppose that for a fixed $A \in M_{3}(\mathbb{R})$, i.e. $3\times 3$ matrix over $\mathbb{R}$, we have linear map $T : M_{3}(\mathbb{R}) \to M_{3}(\mathbb{R})$ defined by $T(B) = AB$ for all $B \in M_{3}(\mathbb{R})$. What is the characteristic polynomial of $T$?
I'm familiar with characteristic polynomials over a matrix, but the extension to a mapping in this fashion is a little odd. May I be steered in the right direction to solve this?
 A: I will use the notation $T_A$ to denote the given operator. I claim that $\chi_{T_A}(x)=\chi_A(x)^3$.
It is easiest to prove this first when $A$ is diagonal, because then the nine eigenvectors of $T_A$ are simply the nine "standard basis matrices" $E_{ij}$ $(1 \leq i,j \leq 3)$ which have $1$ in the $ij^{th}$ place and $0$'s elsewhere. Indeed, when $A=diag(\lambda_1,\lambda_2,\lambda_3)$, then $T_AE_{ij} = \lambda_i E_{ij}$. Since each $\lambda_i$ occurs three times as an eigenvalue of $T_A$, we see that $\chi_{T_A}(x) = (x-\lambda_1)^3(x-\lambda_2)^3(x-\lambda_3)^3 = \chi_A(x)^3$.
Next, we extend the result to the case where $A$ is diagonalizable (not necessarily diagonal). Write $A = U \Lambda U^{-1}$ for some invertible $U$ and diagonal $\Lambda$. Define $P_U: M_3(\Bbb R) \to M_3(\Bbb R)$ by sending $B \mapsto U^{-1}B$. Then we see that $T_A = P_U \circ T_{\Lambda} \circ P_U^{-1}$. Thus we see that $T_A$ is conjugate to $T_{\Lambda}$ in $En d(M_3(\Bbb R))$, and thus $\chi_{T_A}(x) = \chi_{T_{\Lambda}}(x)$. But from the preceding paragraph, we know that $\chi_{T_{\Lambda}}(x) = \chi_A(x)^3$. Hence we again get the result that $\chi_{T_A}(x) = \chi_A(x)^3$.
Finally, we extend this result to nondiagonalizable $A$. The point is that the map $\chi: M_3(\Bbb R) \to \Bbb R^9$ which sends $A$ to the coefficients of the characteristic polynomial of $T_A$, is continuous in the operator norm. Thus by density of diagonalizable matrices in $M_3(\Bbb R)$, it follows that $\chi_{T_A}(x)=\chi_A(x)^3$ for all $A$.
I apologize that this last part of the argument is perhaps non-algebraic and non-elementary. This argument can be generalized to show that if $A \in M_n(\Bbb R)$ then $\chi_{T_A}(x)=\chi_A(x)^n$.
A: The characteristic polynomial for a linear transformation $ T $ on a finite dimensional vector space $ V $, $ T : V \rightarrow V $ is the polynomial $ p( \lambda ) = \det (\lambda I - M) $, where $ M $ is the matrix of $ T $ relative to some ordered basis of $ V $. This is well defined because if $ M_1, M_2 $ are matrices of $ T $ relative to bases $ B_1, B_2 $, then $ M_1 $ and $ M_2 $ are similar.
Here, pick the ordered basis $ \{ E_{ij} \} $, $ 1 \le i,j \le 3 $ for $ M_3( \mathbb{R} ) $, where $ E_{ij} $ denotes the matrix with entry $ 1 $ in the $ i $-th row and $ j $-th column and all other entries zero. Let the columns of $ A $ be $ v_1, v_2, v_3 $. Then we can compute $ T(E_{ij}) = AE_{ij} $ to be the matrix with its $ j $-th column equal to $ v_i $ and the other entries $ =0 $. If $ v_i = (c_{1i}, c_{2i}, c_{3i})^T $, then, $ T(E_{ij}) $ can be written as a sum of basis elements, $ \sum_{k=0}^{3} c_{ki}E_{kj} $. So, for example, the first column of your matrix for $ T $ is then the vector, $ (c_{11}, 0, 0, c_{21}, 0, 0, c_{31}, 0, 0)^T $. Similarly, you can write all of the columns of the matrix $ M $ to get $ M = \{ M_{ij} \} $ where each $ M_{ij} = c_{ij}I_3 $, a $ 3 \times 3 $ block matrix. Now you can compute $ \det (\lambda I - M) $.
Edit as requested: What is $ T(E_{11}) ? $ Let $$ A = \begin{bmatrix} c_{11} & c_{12} & c_{13} \\ c_{21} & c_{22} & c_{23} \\ c_{31} & c_{32} & c_{33} \end{bmatrix} $$ Then we have $$ T(E_{11}) = AE_{11} = \begin{bmatrix} c_{11} & c_{12} & c_{13} \\ c_{21} & c_{22} & c_{23} \\ c_{31} & c_{32} & c_{33} \end{bmatrix} \cdot \begin{bmatrix} 1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} = \begin{bmatrix} c_{11} & 0 & 0 \\ c_{21} & 0 & 0 \\ c_{31} & 0 & 0 \end{bmatrix} $$ and therefore, $$ T(E_{11}) = c_{11}E_{11} + 0E_{12} + 0E_{13} + c_{21}E_{21} + 0E_{22} + 0E_{23} + c_{31}E_{31} + 0E_{32} + 0E_{33} $$ So the co-ordinate vector for $ T(E_{11}) $ in the ordered basis is simply $ (c_{11}, 0, 0, c_{21}, 0, 0, c_{31}, 0, 0)^T $ and therefore this becomes the first column of your matrix.
A: Here's a pretty straightforward way to find the characteristic polynomial of the linear map $T_A(B) = AB$ from $M_n(\Bbb F)$ to itself, where $n \in \Bbb N$ is any natural number and $\Bbb F$ is any field; here I'm disallowing $0 \in \Bbb N$; here my naturals start off with $1$.
So, let's begin by specifying a few things in a little more detail:  first, my readers will have noted that I am admitting vector spaces and matrices over an arbitrary field $\Bbb F$ into this discussion; furthermore, given a square $n \times n$ $B$ matrix with entries selected from $\Bbb F$, we observe that it in fact defines an operator $T_A: M_{n, m}(\Bbb F) \to M_{n, m}(\Bbb F)$, where $M_{n, m}(\Bbb F)$ is the linear space of $n \times m$ matrices over $\Bbb F$; this operator is given by $T_A(B) = AB$ for $B \in M_{n, m}(\Bbb F)$; here we shall in fact find the characteristic polynomial of this linear map for any $m$ in terms of the ordinary characteristic polynimial $\chi_A(x)$ defined as usual in the case $m = 1$; we proceed as follows:
We note that we are here viewing $M_{n, m}$ purely as a linear space over $\Bbb F$; that is, any possible further algebraic structure which $M_{n, m}$ may possess, such as matrix multiplication in the case $m = n$, is simply irrelevant to the problem at hand; for the present concerns, $M_{n, m}(\Bbb F)$ is simply a vector space over $\Bbb F$, and $A$ is a linear map from that space to itself, albeit one with special structural features we may exploit.  Chief amongst these features is seen via the presentation of any $B\in M_{n, m}(\Bbb F)$ in columnar form, thus:
$B = \begin{bmatrix} \mathbf b_1 & \mathbf b_2 & . . . \mathbf b_m \end{bmatrix}, \tag{1}$
where the $\mathbf b_j$, $1 \le j \le m$, are $n \times 1$ column vectors, the columns of $B$.  In terms of this presentation, the action of $T_A$ on $B$ is given by
$T_A(B) = AB = \begin{bmatrix} A\mathbf b_1 & A\mathbf b_2 & . . . A\mathbf b_m \end{bmatrix}; \tag{2}$.
that is, $T_A$ acts on $B$ by operating on each of its columns with the matrix $A$; (2) is easily established using the ordinary definition of matrix multiplication.  Another way of looking at (2) is based upon the fact that $M_{n, m}(\Bbb F)$ is isomorphic, as a vector space over $\Bbb F$, to the direct sum
$\bigoplus_1^m (\Bbb F^n)_j, \tag{3}$
that is, to the direct sum of $m$ copies of $\Bbb F^n$, the "standard" on "cannonical" $n$-dimensional vector space over $\Bbb F$; each $(\Bbb F^n)_j$ may be thought of as the vector space of all possible vectors $ \mathbf b_j \in \Bbb F^n$ which may occupy the $j$-th column of $B$ as $B$ ranges over $M_{n, m}(\Bbb F)$.  In accord with (2), $T_A$ acts upon a vector $\mathbf b$ in the direct sum (3),
$\mathbf b = \bigoplus_1^m \mathbf b_j \in \bigoplus_1^m (\Bbb F^n)_j, \tag{4}$
through the action of $A$ on each summand, jointly and severally:
$T_A(\mathbf b) = T_A(\bigoplus_1^m \mathbf b_j)$
$= \bigoplus_1^m A\mathbf b_j.  \tag{5}$
(3) and (4) are also compatible with the strictly columnar representation if matrices $B \in M_{n, m}(\Bbb F)$; in this form we write the columns of $B$ stacked one above the other, thus:
$[B] = \begin{bmatrix} \mathbf b_1 \\ \mathbf b_2 \\ \vdots \\ \mathbf b_m \end{bmatrix} \in \Bbb F^{nm}. \tag{6}$
The reader familiar with the storage of rectangular arrays in linear computer memory will likely recognize the format (6) for elements of $M_{n, m}(\Bbb F)$ as bearing a certain resemblance to typical software structures often deployed in the management of such data; indeed, if so desired we can write the column (6) as a "flattened" or non-heierarchical  entity by simply re-indexing its entries; we set
$B_{(j - 1)n + k} = (\mathbf b_j)_k, \tag{7}$
where $1 \le j \le m$ and $1 \le k \le n$, to effect such a transformation; we note that the index $(j - 1)n + k$ ranges 'twixt $1$ and $(m - 1)n + n = mn$, exactly as it should to cover the $nm = n \times m$ entries of $B$; the indices of the entries of $\mathbf b_1$ ($j = 1$) range from $1$ to $n$; those of $\mathbf b_2$ ($j = 2$) from $n + 1$ to $2n$, those of $\mathbf b_m$ ($j = m$) from $(m - 1)n + 1$ to $mn$, and so forth.
The action of $T_A$ on the linear space $M_{n, m}(\Bbb F)$ as presented in the form (6), (7) is quite simple to describe; indeed, we may write the matrix $[T_A]$ of $T_A$ as
$[T_A] = \begin{bmatrix} A & 0 & 0 & \ldots & 0 \\ 0 & A & 0 \ldots & 0 \\ 0 & 0 & A & \ldots & 0 \\ \vdots \\ 0 & 0 & 0 & \ldots & A \end{bmatrix}, \tag{8}$
i.e., $[T_A]$ is a square block-diagonal matrix of size $mn$, the $m$ diagonal blocks of which are each equal to the size $n$ square matrix $A$.  In accord with (5), (6) we then have, in this format, 
$T_A(B) = [T_A][B] =$
$= \begin{bmatrix} A \mathbf b_1 \\ A \mathbf b_2 \\ \vdots \\ A \mathbf b_m \end{bmatrix}; \tag{9}$
the eigen-equation for $T_A$ may thus be expressed:
$[T_A][B] = \begin{bmatrix} A & 0 & 0 & \ldots & 0 \\ 0 & A & 0 \ldots & 0 \\ 0 & 0 & A & \ldots & 0 \\ \vdots \\ 0 & 0 & 0 & \ldots & A \end{bmatrix} \begin{bmatrix} \mathbf b_1 \\ \mathbf b_2 \\ \vdots \\ \mathbf b_m \end{bmatrix} = \lambda \begin{bmatrix} \mathbf b_1 \\ \mathbf b_2 \\ \vdots \\ \mathbf b_m \end{bmatrix}$
$= \begin{bmatrix} \lambda \mathbf b_1 \\ \lambda \mathbf b_2 \\ \vdots \\ \lambda \mathbf b_m \end{bmatrix}, \tag{10}$
which may also be written
$([T_A] - \lambda I_{mn})[B] = 0, \tag{11}$
where $I_{mn}$ is the $mn \times mn$ identity matrix.  The characteristic polynomial of $[T_A]$, and hence of $T_A$, is thus
$\chi_{T_A}(\lambda) = \det([T_A] - \lambda I_{mn}); \tag{12}$
we see that
$[T_A] - \lambda I_{mn} = \begin{bmatrix} A - \lambda I_n & 0 & 0 & \ldots & 0 \\ 0 & A - \lambda I_n & 0 \ldots & 0 \\ 0 & 0 & A- \lambda I_n & \ldots & 0 \\ \vdots \\ 0 & 0 & 0 & \ldots & A- \lambda I_n \end{bmatrix}, \tag{13}$
and thus that
$\chi_{T_A}(\lambda) = \det(\begin{bmatrix} A - \lambda I_n & 0 & 0 & \ldots & 0 \\ 0 & A - \lambda I_n & 0 \ldots & 0 \\ 0 & 0 & A- \lambda I_n & \ldots & 0 \\ \vdots \\ 0 & 0 & 0 & \ldots & A- \lambda I_n \end{bmatrix}). \tag{14}$
But the determinant of a block diagonal matrix is the product of the determinants of its diagonal blocks, see https://en.m.wikipedia.org/wiki/Block_matrix; from this fact, and from (14), we find that
$\chi_{T_A}(\lambda) = (\det(A - \lambda I_n))^m = (\chi_A(\lambda))^m, \tag{15}$
since $[T_A] - \lambda I_{mn}$ consists of $m$ diagonal blocks each of the form $A - \lambda I_n$.  In the specific problem at hand, $m= n$ and thus
$\chi_{T_A}(\lambda) = (\chi_A(\lambda))^n, \tag{16}$
which holds over any field $\Bbb F$, in particular over $\Bbb R$; finally, since $n = 3$ here, $A \in  
M_3(\Bbb R)$ and
$\chi_{T_A}(\lambda) = (\chi_A(\lambda))^3. \tag{17}$
In closing, I would like to point out that the preceding argument not only generalizes the stated question to the cases $m \ne n$ and arbitrary fields $\Bbb F$, but is also in fact purely algebraic in nature, and does not rely on continuity or other topological concepts as does the answer given by Shalop, though I feel Shalop's answer is nice work and in fact gave it an upvote.  
