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Let $\text{Mat}_n(\mathbb{C})$ be the set of $n \times n$ complex matrices. Let $\sigma\colon \text{Mat}_n(\mathbb{C}) \rightarrow \text{Mat}_n(\mathbb{C})$ be the map that takes a matrix to its semisimple part. In more detail, consider $M \in \text{Mat}_n(\mathbb{C})$. Let $V_1,\ldots,V_k$ be the generalized eigenspaces of $M$, and for $1 \leq i \leq k$ let $\lambda_i$ be the generalized eigenvalue associated to $V_i$. Thus $\lambda_1,\ldots,\lambda_k$ are the roots of the characteristic polynomial of $M$, the subspace $V_i$ is the kernel of $(M - \lambda_i)^n$, and $\mathbb{C}^n$ is the direct sum of the $V_i$. The image $\sigma(M)$ is then the matrix whose associated linear map scales each $V_i$ by $\lambda_i$.

Question: Is $\sigma$ an algebraic map? Or at least holomorphic?

If I had seen this question a few days ago, my impulse would be that the answer has to be "yes". However, as a byproduct of a proof of an unrelated fact, I have what looks like a very indirect proof that the answer is "no". I'm asking for a direct proof as a kind of "idiot test" to see if my reasoning is flawed.

Actually, it looks like a bit of a pain to even see that $\sigma$ is continuous, though surely that must be true!

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    $\begingroup$ If I understand your description correctly, then $\sigma$ is the identity on diagonalizable matrices with no multiple eigenvalues. These are dense, so $\sigma$ is not continuous. $\endgroup$
    – Christian Remling
    Commented Aug 2, 2017 at 0:10
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    $\begingroup$ @ChristianRemling: Oh, good point! That pretty much kills the question. I'll delete it (after waiting a while to make sure you can see my thanks!). $\endgroup$
    – Helen
    Commented Aug 2, 2017 at 0:27
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    $\begingroup$ Actually, it doesn't look like I can delete it, so I suppose it stays! $\endgroup$
    – Helen
    Commented Aug 2, 2017 at 0:42
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    $\begingroup$ Another way of seeing @ChristianRemling's point: the matrices $\begin{pmatrix} \varepsilon & 1 \\ 0 & 0 \end{pmatrix}$ are semisimple for $\varepsilon \ne 0$, but nilpotent for $\varepsilon = 0$. I remember that, despite the simplicity (no pun intended) of this example, I (like you) was so convinced that $\sigma$ must be continuous that I spent some time trying to find a mistake in it. $\endgroup$
    – LSpice
    Commented Aug 2, 2017 at 2:05
  • $\begingroup$ @LSpice: Cool, thanks! That's an enlightening example. $\endgroup$
    – Helen
    Commented Aug 2, 2017 at 2:23

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