# Eigenvalues of a linear operator on $\mathbb{C}^{m \times n}$

There is a lemma in the second addition of Artin's Algebra (Chapter 10 on Representation Theory) used to prove the orthogonality relations for irreducible characters:

$\textbf{Lemma :}$ Let $T$ be an operator on the space $\mathbb{C}^{m \times n}$ of $m \times n$ complex matrices, defined as

$T(M) = AMB$, where $A$ is some $m \times m$ complex matrix and $B$ some $n \times n$ complex matrix.

Then trace $T =$ trace$(A)$trace$(B)$.

One can construct an "eigenmatrix" (Is this terminology standard?) of $T$ from an eigenvector $x$ of $A$ and $y$ of $B^T$, namely the matrix $xy^T$. Consequently if $x$ was an eigenvector of eigenvalue $\lambda$ and $y$ of eigenvalue $k$, then the eigevalue of the eigenmatrix $xy^T$ is $\lambda k$.

Now to find the trace of $T$, one needs all its eigenvalues. A sufficient but perhaps not necessary condition would be if $T$ has $mn$ distinct eigenvalues. If it does then the lemma in Artin above follows immediately.

However the question that arises is if $T$ does not have $mn$ distinct eigenvalues. Artin does mention of constructing a sequence of matrices $A_k \rightarrow A$ and $B_k \rightarrow B$ such that each of these have distinct eigenvalues and the product of their eigenvalues is distinct for all $k$. He the proceeds to say that the lemma follows by continuity.

I find this argument not rigorous - how can one choose a sequence matrices $A_k$ and $B_k$ like that? Can it made to be rigorous? Someone said to me that one can prove this lemma by using Jordan normal form. Does anyone know of a proof using Jordan form?

$\textbf{ Update :}$ This lemma is more easily proven by bashing out the algebra as shown in Robert and Hans's answers below. However, how about if we were to ask of computing the eigenvalues of $T$?

Thanks.

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Not that this really answers your questions, but why not simply prove the lemma by computation?

If $E^{(ij)}$ is the matrix with a $1$ at position $(i,j)$ and zeros elsewhere, the "$(ab,ij)$" matrix entry when representing the operator $T$ in the basis $\{ E^{(ij)}\}$ is $$(A E^{(ij)} B)_{ab} = \sum_{k,l} A_{ak} E^{(ij)}_{kl} B_{lb} = A_{ai} B_{jb}.$$ Summing the diagonal entries "$(ij,ij)$" gives $$\operatorname{Tr} T = \sum_{i,j} (AE^{(ij)}B)_{ij} = \sum_{i,j} A_{ii} B_{jj} = \left( \sum_i A_{ii} \right) \left( \sum_j B_{jj} \right) = \operatorname{Tr} A \operatorname{Tr} B.$$

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I realise now that computing the trace would have been just easier by bashing the algebra straight out. However in Artin he tries to compute the trace of $T$ by first working out the eigenvalues. If one wanted to do it this way, I am trying to figure that out now. – user38268 Oct 19 '11 at 9:19
Do you know how can one compute all* the eigenvalues of such an operator? I have edited my question above. Thanks. – user38268 Oct 19 '11 at 20:26
@Benjamin: I don't think I have anything sensible to add to what you've already said in the question, except perhaps that matrices with distinct eigenvalues are dense in the space of all square matrices of a given size (since eigenvalues coincide iff the discriminant vanishes; this condition defines a hypersurface in the space, and the complement of a hypersurface is dense). That should be the crucial ingredient in making the approximation argument rigorous. – Hans Lundmark Oct 20 '11 at 5:52
Thanks for sharing what you know, I will see what I can try to understand. – user38268 Oct 20 '11 at 6:12
Dear Hans, do you mean that "... since eigenvalues coincide iff the determinant vanishes ..." instead of "... since eigenvalues coincide iff the discriminant vanishes ...". Regards, – Amitesh Datta Oct 20 '11 at 11:30

Yes, it is easy to do using Jordan normal form, since for a matrix in Jordan form it is easy to perturb the eigenvalues by changing the diagonal elements slightly. But Jordan form is not available over every field, and I think this is easier to prove more directly. Note that both sides of the equation $\text{trace }T = (\text{trace }A)(\text{trace }B)$ are linear in $A$ and in $B$, so it suffices to prove for $A$ and $B$ in a suitable basis. So we can take $A = e_i e_j^T$ and $B = e_k e_l^T$ for $1 \le i,j \le m$ and $1 \le k,l \le n$, where $e_i$ is the vector (of appropriate size) with 1 in the $i$'th place and 0 everywhere else. For ${\mathbb C}^{m \times n}$ use the basis of elements $e_r e_s^T$. Then $$\text{trace} T = \sum_{r,s} ((e_i e_j^T e_r e_s^T e_k e_l^T))_{r,s} = \sum_{r,s} \delta_{ir} \delta_{jr} \delta_{sk} \delta_{ls} = \delta_{ij} \delta_{kl} = \text{trace}(e_i e_j^T) \text{trace}(e_k e_l^T)$$

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I see what you mean immediately if we ask about the trace. It is indeed easier to calculate the trace by bashing the algebra straight away than to calculate the eigenvalues. However what if now I would like to calculate all the eigenvalues of $T$ (And then use this result to calculate the trace) ? – user38268 Oct 19 '11 at 9:18
Do you how one can compute all the eigenvalues of such an operator $T$? Hans above has given an argument but I am not well versed in such techniques. – user38268 Oct 20 '11 at 6:13