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Which of the following are compact?

$1$. The set of all upper triangular matrices all of whose eigenvalues satisfy $|\lambda|\leq 2$.

$2$. The set of all real symmetric matrices all of whose eigenvalues satisfy $|\lambda|\leq 2$.

$3$. The set of all diagonalizable matrices all of whose eigenvalues satisfy $|\lambda|\leq 2$.

Intuitively I feel that all these are bounded from the condition. Is it true? For $1 $and $ 2$ I think the set is closed. $3$ I am not sure. Please give a detailed answer I have never been exposed to these problems before. Thanks

copper.hat
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  • What do open covers of these sets look like? I'm not sure if this proves $(1)$ is not compact, but I believe that the set $$\left{ \left[\begin{array} \1&n\0& 1 \end{array}\right]\right}_{n=1}^\infty$$ is a subset of all upper triangular matrices with $|\lambda| \leq 2$ – graydad Nov 28 '14 at 17:25
  • No, they are not all bounded. – copper.hat Nov 28 '14 at 17:25
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    @graydad: That proves that it is not compact as those matrices are unbounded. Similarly, $\left{ \left[\begin{array} \1&n\0& 2 \end{array}\right]\right}_{n=1}^\infty$ shows that (3) is not compact. You can see (2) in various ways, one way is to note that such matrices are unitarily diagonalizable. – copper.hat Nov 28 '14 at 17:26
  • @graydad no, becase your matrices have the trace $n+1/n$, which is unbounded, yet the sum of eigenvalues must be bounded by $4$ in absolute value, so they are not in the set described in (2). – TZakrevskiy Nov 28 '14 at 17:44
  • Given that (1) and (3) are already there, here's the hint for (2): the matrices are a finite-dimensioned vector space, so all norms are equivalent. Take the spectral norm $|A|_2$, which, for $A=A^*$ is equal to the spectral radius of $A$. – TZakrevskiy Nov 28 '14 at 17:46
  • @graydad there were some misunderstandings in my previous comment, I edited them out. Still, your set of matrices is not a subset of the set described in (2). – TZakrevskiy Nov 28 '14 at 17:49
  • @graydad yes, but the other eigenvalue is equal to $n+1/n> 2$ for $n>1$, which is not allowed in (2). We need all eigenvalues to be bounded by $|\lambda|\le 2$. – TZakrevskiy Nov 28 '14 at 17:53

1 Answers1

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Let $A_n = \begin{bmatrix}1&n\\0& 2 \end{bmatrix}$, we see that the eigenvalues of $A_n$ are $1,2$ and $A_n$ is upper triangular. Since the eigenvalues are distinct, $A_n$ is diagonalisable. We have $\|A_n e_2\|_2 = \sqrt{n^2+4}$, hence $\|A_n\|_2 \ge n$ and so the $A_n$ are unbounded. It follows that the sets (1), (3) are unbounded, hence are not compact.

The real symmetric matrices are normal and so can be diagonalised by an orthogonal matrix. The set of orthogonal matrices $O(n)$ is compact (closed since the map $U \mapsto U U^T-I$ is continuous, and bounded since $\|U\|_2 = 1$), and the set ${\cal D}$ of diagonal matrices whose diagonal entries lie in $[-2,2]$ is compact. Let ${\cal S}$ denote the $n \times n$ real symmetric matrices whose eigenvalues lie in $[-2,2]$. Since $\phi:O(n) \times {\cal D} \to {\cal S}$ given by $\phi(U,\Lambda) = U \lambda U^T$ is continuous, and $\phi$ is surjective, it follows that ${\cal S}$ is compact.

copper.hat
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  • I think the results (1) & (3) can be generalized. That means if we omit the condition $|\lambda|\le 2$ then the sets are also NOT compact.Otherwise if we impose the condition $|\lambda|\le k$ for some $k\in \mathbb R$ then the sets are again NOT compact.Am I right ? or wrong? Please help on my argument. @copper.hat – Empty Feb 15 '15 at 06:18
  • Is the condition $|\lambda|\le 2$ necessary to prove the option (2)? Can it be generalized as $|\lambda|\le k$ for some $k\in \mathbb R$ ? @copper.hat. – Empty Feb 15 '15 at 06:33
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    Yes. The map $f(A) = {k \over 2} A$ is continuous, invertible and maps the set of symmetric matrices whose eigenvalues satisfy $|\lambda| \le 2$ to the set of symmetric matrices whose eigenvalues satisfy $|\lambda| \le k$. – copper.hat Feb 15 '15 at 07:10
  • So the condition $|\lambda|\le 2$ is sufficient, NOT necessary. – Empty Feb 15 '15 at 07:12
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    Of course. The question is asking if the given sets are compact. – copper.hat Feb 15 '15 at 07:16