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Is the set of all diagonalizable matrices compact?

I have got this example \begin{bmatrix} 1 && n \\ 0 && 2\end{bmatrix} which is diagonalizable but not compact because it is not bounded.Also all its eigen values are bounded.

But I don't know why one writes that

A real symmetric matrix whose eigenvalues are bounded is compact

  • The justification given is that it is similar to a diagonal matrix

Am I missing something?

Learnmore
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3 Answers3

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The diagonalizable matrices are not bounded, hence not compact. If you enforce a bound on the spectral radius or on some norm of choice (a natural thing to do, given the previous remark), then they are still not compact, because they are not closed, as you can see through examples like:

$$A_n = \begin{bmatrix} 1 & 1+1/n \\ 1/n & 1 \end{bmatrix}.$$

This has trace $2$ and determinant $1-1/n-1/n^2$, which is not $1$, so it has distinct eigenvalues, hence it is diagonalizable. But $\lim_{n \to \infty} A_n$ is one of the classic examples of a nondiagonalizable matrix.

More can be said: the set $D$ of all diagonalizable matrices of a given size turns out to be dense in the set $M$ of all the square matrices of a given size. So $\overline{D} = M$, which is considerably larger than $D$, so $D$ is not closed. But this is harder to prove than my simple example.

Ian
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  • Even if the spectral radius is bounded the matrix can still be diagonalisable but not bounded – Learnmore Jan 21 '16 at 06:34
  • you have said they are not closed they are not bounded also ;right?? – Learnmore Jan 21 '16 at 06:35
  • thus it means the set of all real symmetric matrices with bounded spectral radius is NOT compact;right – Learnmore Jan 21 '16 at 06:38
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    @Amartya You are correct that just enforcing a bound on the spectral radius doesn't give a bounded set because of examples like $\begin{bmatrix} 1 & M \ 0 & 1 \end{bmatrix}$ for large $M$. However, this isn't true in the symmetric matrices, for which the spectral radius is a norm (indeed it coincides with the Euclidean operator norm). – Ian Jan 21 '16 at 13:57
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The set $S:=\{A\in M_n(\mathbb R): A=A^\top\land\rho(A)\le K\}$ where $K$ is a constant, is compact. See this question which tries to prove a similar similar result. $\rho$ denotes spectral radius.

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Jack's wasted life
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Your wording "A real symmetric matrix whose eigenvalues are bounded is compact" does not make sense.

For any given real/complex matrix (as far as it is a square matrix of finite size) the set of eigenvalues is finite subset of the set of complex numbers. And any finite set is a bounded set.

SLightly digressing: The set of diagonalizable complex matrices is a (Zariski) dense subset of the space of all complex matrices of that size.

P Vanchinathan
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