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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!

Noble.
  • 2,446

3 Answers3

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

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

Sugata Adhya
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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 $$ \tr L_A = \tr(A \otimes I_n) = \tr(A)\tr(I_n) = n\tr(A),\\ \det L_A = \det(A \otimes I_n) = \det(A)^n\det(I_n)^n = \det(A)^n. $$

Branimir Ćaćić
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1

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 $\det(T) = \det(I\otimes A) = \det(A)^m$.

user1551
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  • How does one find the determinant and trace using the approach of the first answer? I understand that we are supposed to find the eigenvalues for this, but how do we do that? I also don't understand the $\otimes$ notation in other answers... – user2154420 Nov 03 '14 at 23:44
  • @user2154420 The symbol $\otimes$ here denotes a Hadamard product of two matrices. In the special case $I\otimes A$, the Hadamard product happens to be a block diagonal matrix of size $m^2\times m^2$, where each diagonal block is equal to $A$. Therefore the trace and determinant follow immediately and you don't need to calculate any eigenvalue. – user1551 Nov 04 '14 at 00:48
  • I understand how you calculate the trace, but how do you calculate the determinant from that? – user2154420 Nov 04 '14 at 01:05
  • @user2154420 The determinant of a block diagonal matrix (or more generally, that of a block triangular matrix) is the product of the determinant of the diagonal subblocks. – user1551 Nov 04 '14 at 01:23
  • I still don't understand... our matrix overall isn't block diagonal or triangular. I'm assuming our blocks are $a_{i j}I_m$? – user2154420 Nov 05 '14 at 01:45
  • @user2154420 $I\otimes A$ is always a block-diagonal matrix, where each diagonal block here is equal to $A$ (see my comment above). That is, you have $m$ copies of $A$ along the diagonal, and zeros elsewhere. In Branimir Ćaćić's answer, the $(i,j)$-th block does equal to $a_{ij}I_m$, but this is not the case here. My matrix is $I\otimes A$, which is block diagonal, whereas his is $A\otimes I$, which isn't block diagonal. The difference arises because we order the basis vectors differently. – user1551 Nov 05 '14 at 03:23
  • So is there any easy way to take the determinant from his matrix? – user2154420 Nov 06 '14 at 00:52
  • Oh, nevermind. I see how you get the diagonal matrix... you order your column vector representation of $B$ differently. – user2154420 Nov 06 '14 at 01:14