# Problem on the compactness of a subset of $\mathsf{M}_n(\mathbb{R})$

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 ???

• 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 $A$). Jul 24, 2019 at 6:04

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