Is the set of all real symmetric matrices all of whose eigenvalues satisfy $|λ| \le 2 $ compact?

Question

Is the set of all real symmetric matrices all of whose eigenvalues satisfy $|λ| \le 2 $ compact?

Let $S$ be the set of all real symmetric matrices all of whose eigenvalues satisfy $|λ| \le 2 $.

Define $f:M_n(\Bbb R)\to M_n(\Bbb R)$ by $f(A)=A-A^T$ which is continuous .

Then $S=f^{-1}(0)$ and hence is closed.

Also any $A\in S$ is diagonalisable and hence $A=PDP^T$ where $D$ is a diagonal matrix and its diagonal entries are the eigen values of $A$.

But the diagonal entries of $D$ are $\le 2$ and hence bounded.

How can we conclude that $S$ is bounded from here?

Please help.

Answer 1

Such matrices are $U^TDU$ where $U\in O(n)$ and $D\in \mathcal A$ where $\mathcal A$ is the set of diagonal matrices with entries in $[-2,2]$. Both $O(n)$ and $\mathcal A$ are compact sets. The set of your matrices is the image of a compact set under a continuous map, so compact.

Answer 2

No, you cannot deduce that $S$ is bounded from what you did. Did you notice that at no moment you used the fact that your matrices are symmetric? And, without that, $S$ contains, for instance, the set of all strictly upper triangular matrices, which is clearly unbounded.

Answer 3

YES.

First of all, $\mathcal A$ is bounded with respect to the induced $2-$norm $$ \|A\|=\sup_{\|x\|=1}\|Ax\|, $$ since, if $A$ is symmetric (or more generally, normal) then $$ \|A\|=\max\{|\lambda|: \lambda \,\,\,\text{eigenvalue of $A$}\}, $$ and hence is our case $$ \|A\|\le 2 \quad\text{for all}\quad A\in\mathcal A. $$ In $\{A_k\}\subset \mathcal A$, is a convergent sequence, with $A_k\to A\in\mathcal M_n(\mathbb R)$, then $A$ is symmetric, since $$ \|A^T-A\|= \|A^T-A^T_k+A_k-A\|\le \|A^T-A^T_k\|+\|A_k-A\|\to 0, $$ as $k\to \infty$, and hence $A^T=A$, and also $$ \big|\|A_k\|-\|A\|\big|\le \|A_k-A\|\to 0, $$ and hence $$ \max\{|\lambda|: \lambda \,\,\,\text{eigenvalue of $A$}=\|A\|=\lim_{k\to\infty}\|A_k\|\le 2. $$ Thus $A\in\mathcal A$, and hence $\mathcal A$ is closed and bounded.

Answer 4

Being symmetric is a closed condition, having only eigenvalues $\in[-2,2]$ is a closed condition. So all that's left is boundedness. For any such matrix $A$, we have $|x^TAx|\le 2|x|^2$, hence by taking $x=e_i$, $|A_{ii}|\le 2$, and by taking $x=e_i+e_j$ and using that $A$ is symmetric(!), $|A_{ii}+A_{jj}+2A_{ij}|\le 2$, so $|A_{ij}|\le 3$.