Consider $$X=\Big\{A \in M_3(\Bbb R): \rho_A(x)=x^3-3x^2+2x-1\Big\}$$ where $\rho_A(x)$ is the characteristic polynomial of $A$ and $M_3(\Bbb R)$ is the space of all $3 \times 3$ matrices over $\Bbb R$.

Is $X$ compact in $M_3(\Bbb R)$ ?

My try: I confused with only the $2x$ term in $\rho_A(x)$. Because , if the $2x$ term not appear , then $X$ becomes $$X=\{A \in M_3(\Bbb R): \text{trace}(A)=3,\det A=1\} $$ which is unbounded, since $$(\forall n \in \Bbb N):\begin{pmatrix} 1&0 & n\\0&1&0\\0&0&1 \end{pmatrix} \in X$$

But here the problem is, the appearance of $2$. I Know $$2=A_{11}+A_{22}+A_{33}$$

so I think in this case the set becomes bounded and closed

Any help?

  • $\begingroup$ I hope my earlier incorrect comment did not mislead you. The characteristic polynomial has three distinct eigenvalues, so any $A \in X$ must have the same eigenvalues. $\endgroup$ – copper.hat Dec 10 '18 at 3:00
  • $\begingroup$ If you were dealing with complex elements then the set $X$ would be unbounded (take an upper triangular matrix). $\endgroup$ – copper.hat Dec 10 '18 at 3:13

For example, $$ \pmatrix{t & 0 & 1\cr 1-2t & 0 & -2\cr -t^2+3t & 1 & 3-t\cr} $$ has that characteristic polynomial. This is $S^{-1} A S$ where $$ A = \pmatrix{0 & 0 & 1\cr 1 & 0 & -2\cr 0 & 1 & 3\cr},\ S = \pmatrix{1 & 0 & 0\cr 0 & 1 & 0\cr t & 0 & 1\cr}$$

  • $\begingroup$ Can you shed light on how this very nice answer might have been guessed -- and whether specifying the characteristic polynomial (or some annihilating polynomial) ever yields a compact set? $\endgroup$ – user25959 Dec 10 '18 at 3:19
  • $\begingroup$ @Robert Israel: super! thanks! what is the intuition behind for finding this matrix ? $\endgroup$ – Chinnapparaj R Dec 10 '18 at 3:21
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    $\begingroup$ Start with a companion matrix for that characteristic polynomial. Choose an $S$ such that $S$ and $S^{-1}$ have entries that are polynomial in $t$... $\endgroup$ – Robert Israel Dec 10 '18 at 3:22
  • $\begingroup$ Of course there's the trivial case where the minimal polynomial has degree $1$, so the matrix is a constant multiple of $I$. $\endgroup$ – Robert Israel Dec 10 '18 at 3:25
  • $\begingroup$ Thank you sir! :-) $\endgroup$ – Chinnapparaj R Dec 10 '18 at 3:28

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