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Prove or disprove : if $S$ is the set of all $3\times 3$ square matrices $A$ where characteristic polynomial of $A$ is $\chi_A = X^3-3X^2+2X-1$, then $S$ compact, ($S$ topologized by $\mathbb{R}^9$).

Actually, I am just new to the concepts of characteristic polynomial and eigenvalues of a matrix but here $\operatorname{tr}(A) = -3$ and $\det(A) = -1$, and there is only one real eigenvalue of $A$, for all $A$ in $S$. But, I can't determine other facts about those matrices A, hence I can't think about closed and boundedness of $S$, but I think S will be unbounded. Can someone refer me good books for good problems in matrix topology ???

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    $\begingroup$ The fact that $S$ is closed should follow from the continuity of the map that takes $A$ to its characteristic polynomial (that is, each coefficient of the characteristic polynomial is a continuous function of the nine entries of $A$). $\endgroup$ Jul 24, 2019 at 6:04

2 Answers 2

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Here is a brute force approach:

The eigenvalues of $A$ are $r, a\pm ib$. Let $L=\begin{bmatrix} r & 0 & 0 \\ 0 & a & -b \\ 0 & b & a \end{bmatrix}$, and $V_n=\begin{bmatrix} 1 & 0 & n \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$, then $V_n^{-1} = V_{-n}$.

Then $[V_n L V_n^{-1}]_{12} = -nb$.

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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.

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