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From Artin's Algebra (2nd edition)

Exercise 5.2.3 Let $A$ be an $n\times n$ complex matrix.

(a) Consider the linear operator $T$ defined on the space $\mathbb{C}^{n\times n}$ of all complex $n\times n$ matrices by the rule $T(M)=AM-MA$. Prove that the rank of this operator is at most $n^2-n$.

(b) Determine the eigenvalues of $T$ in terms of the eigenvalues $\lambda_1\cdots\lambda_n$ of $A$.

This is an exercise in the section "Using Continuity", which introduces the proof of Cayley-Hamilton theorem via first proving the theorem for diagonalizable matrices then extending this result to all matrices via a continuity argument. I have tried to replicate this technique for the problem above. I have succeeded in showing the result in (a) for diagonalizable matrices, but I'm not sure how to proceed using continuity.

Here's my idea: Let $A_i\to A$ be a sequence of matrices, with limit $A$ and each having distinct eigenvalues. The rank of $T$ for each $A_i$ (denote it $T_i$) is at most $n^2-n$. If $X_i\in\ker T_i$ and $X_i\to X$, then $$T(X) = \lim_{I \to \infty} T_i(X_i) = 0$$ by continuity. Intuitively, this sequence of subspaces of dimension $\le n^2-n$ should "approach" some subspace of dimension $\le n^2-n$, but I'm not sure how to formalize this.

Rodrigo de Azevedo
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pzq_alex
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  • Your idea is a fine start. $T$ is a finite dim linear map so it has some matrix representation... what do you know about the minors (sub-matrix determinants) of $T_i$, and what does that imply for the minors of $T$ itself? – user8675309 May 26 '23 at 16:25
  • I think that link misses the easiest explanation for the defective matrix case, which is that that polynomial functions are continuous so $\text{rank }T_i \leq r$ for all $i \implies$ for $k\in \big{1,2,...,n-r\big}$ that all $r+k \times r+k$ minors are zero for all $i$ hence the limit is zero as well, i.e. $\text{rank }T \leq r$. – user8675309 May 27 '23 at 15:48

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