Finding trace and determinant of linear operator

Question

I've got the following question

Consider the linear operator of left multiplication by an $m \times m$ matrix $A$ on the vector space of all $m \times m$ matrices. Determine the trace and determinant of this operator.

I'm a bit stuck as to how to even begin, I know this is going to involve eigenvalues/vectors and that if $\lambda_1, \lambda_2, ... ,\lambda_m$ are the $m$ roots of the characteristic polynomial of an $m \times m$ matrix $A$, then:

$\det(A) = \lambda_1 ... \lambda_m$

and

$\text{trace}(A) = \lambda_1 + ... + \lambda_m$

But obviously not all $m \times m$ matrices have $m$ eigenvalues so I'm really stuck on this question.

Thanks!

Answer 1

$$T:Mat_{m\times m}(F)_F\to Mat_{m\times m}(F)_F:B\mapsto AB$$

$\{E_{ij}=(e_{pq})\in Mat_{m\times m}(F):i,j=1,2,...,m\}$ is a basis of $Mat_{m\times m}(F)$ where

$$e_{pq}=\begin{cases}1\text{ if $i=p,j=q$ ;}\\0\text{ otherwise.}\end{cases}$$

It's a matter of verification that the $i^{th}$ column of $A$ becomes the $j^{th}$ column of

$T(E_{ij})=AE_{ij}.$ Take $A=(a_1, a_2,...,a_m)$ column-wise. Then

$$T(E_{11})=(a_1,0,0,...,0)=a_{11}E_{11}+a_{21}E_{21}+...+a_{m1}E_{m1}\\T(E_{12})=(0,a_1,0,...,0)=a_{11}E_{12}+a_{21}E_{22}+...+a_{m1}E_{m2}\\...\\T(E_{1m})=(0,0,0,...,a_1)=~...\\T(E_{21})=(a_2,0,0,...,0)=~...\\...\\T(E_{2m})=(0,0,0,...,a_2)=~...\\...\\...\\T(E_{m1})=(a_m,0,0,...,0)=~...\\...\\T(E_{mm})=(0,0,0,...,a_m)=~...\\$$

Therefore (verify) $$T=\begin{pmatrix}a_{11}I_{}&a_{12}I&...&a_{1m}I\\a_{21}I&a_{22}I&..&a_{2m}I\\...&...&...&...\\a_{m1}I&a_{m2}I&...&a_{mm}I\end{pmatrix}$$

where $I$ is the identity matrix of order $m.$

Answer 2

$\DeclareMathOperator{tr}{Tr}$ Let $\tau : M_n(F) \cong F^n \otimes_F F^n$ be the isomorphism given by $\tau : S \mapsto \sum_{i,j=1}^n S_{ij} e_i \otimes e_j$. Then $$ \tau(L_A S) = \sum_{i,j=1}^n (AS)_{ij} e_i \otimes e_j = \sum_{i,j=1}^n \left(\sum_{k=1}^n A_{ik}S_{kj}\right)e_i \otimes e_j = \sum_{k,j=1}^n S_{kj}\left(\sum_{i=1}^n A_{ik}e_i\right) \otimes e_j\\ = \sum_{k,j=1}^n S_{kj} \left(Ae_k \otimes e_j\right) = (A \otimes I_n) \tau(S), $$ so that if $L_A \in L(M_n(F))$ denotes left multiplication by $A$, then $$ \tr L_A = \tr(A \otimes I_n) = \tr(A)\tr(I_n) = n\tr(A),\\ \det L_A = \det(A \otimes I_n) = \det(A)^n\det(I_n)^n = \det(A)^n. $$

Answer 3

The linear transformation is $T:X\mapsto AX$, i.e. $\operatorname{vec}(X)\mapsto (I\otimes A)\operatorname{vec}(X)$. Hence $\operatorname{tr}(T) = \operatorname{tr}(I\otimes A) = m\operatorname{tr}(A)$ and $\det(T) = \det(I\otimes A) = \det(A)^m$.