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!
Best Answer
$$T:Mat_{m\times m}(F)_F\to Mat_{m\times m}(F)_F:B\mapsto AB$$
$\{E_{ij}=(e_{pq})\in Mat_{m\times m}(F):i,j=1,2,...,m\}$ 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 $m.$