Whether the set $A$ is a compact subset of $M_3(\Bbb R)$.

Question

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 $M_3(\Bbb R)$ ?

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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?

Answer 1

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 $$ A = \pmatrix{0 & 0 & 1\cr 1 & 0 & -2\cr 0 & 1 & 3\cr},\ S = \pmatrix{1 & 0 & 0\cr 0 & 1 & 0\cr t & 0 & 1\cr}$$

Answer 2

First, observe that the given polynomial has a real root call is $\alpha$. Now the other two roots are both real or complex. If it is complex call it $\omega, \overline{\omega}$. Consider the matrix $A_n= \begin{pmatrix} \alpha& 0 & n \\ 0 & 0 & |\omega|^2 \\ 0& -1& 2Re \omega\\ \end{pmatrix};$ $n \in \mathbb{N}$. It's characteristic polynomial is given polynomial. If both the root are real call it $ \beta, \gamma$, then consider the matrix$ A'_n=\begin{pmatrix} \alpha& 0 & n \\ 0 & \beta & 0\\ 0& 0& \gamma\\ \end{pmatrix};$ $n \in \mathbb{N}$. We get an unbounded set as $\|A_n\|\geq n$ and $\|A'_n\| \geq n$ when $ M_3(\mathbb{R})$ is identified with $\mathbb{R}^9$. I think there is nothing special about that polynomial.