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$\mathbb{D}(n;\mathbb R)$ be set of all $n \times n$ diagonalizable matrices over $\mathbb R$. Which of the following are true?

(a) Subspace of $\mathbb{M}(n;\mathbb R)$

(b) Connected

(c) Compact

(d) Dense

(e) None of the above

(a) $O$ matrix is in $\mathbb{D}(n;\mathbb R)$. $A \in\mathbb{D}(n;\mathbb R)$, then $kA \in \mathbb{D}(n;\mathbb R)$ $\forall\alpha \in \mathbb{R}$. I am not able to prove the statement 'suppose $A,B \in \mathbb{D}(n;\mathbb R)$ then $A+B \in \mathbb{D}(n;\mathbb R)'$.

(b)If I could prove it is a subspace, I can prove it is path connected, hence it is a connected space in the set of all matrices.

(c)I don't think it is bounded with respect to Euclidean norm in $\mathbb{R^{n^2}}$. Hence it is not compact.

(d) I don't know how to approximate any matrices with elements of $\mathbb{D}(n;\mathbb R)$

Please help me to judge the answer.

1 Answers1

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a) $\begin{pmatrix}1&0\\0&0 \end{pmatrix}$ and $\begin{pmatrix}0&1\\0&1 \end{pmatrix}$ are both diagonalizable but their sum isn't.

b) $\mathbb{D}(n;\mathbb R)$ is indeed path-connected: each element has a path to the identity matrix: if $A=PDP^{-1}$, consider $\gamma:t\mapsto P((1-t)D+tI)P^{-1}$.

c) $\mathbb{D}(n;\mathbb R)$ is clearly unbounded as you noticed.

d)Consider the mapping $f:\begin{pmatrix}a&b\\c&d \end{pmatrix} \mapsto (a-d)^2+4bc$.

$f(M)$ is the discriminant of the characteristic polynomial of $M$. When $M$ is diagonalizable, $f(M)\geq 0$. There are matrices with no real eigenvalues. For these matrices, $f$ is $<0$. Since $f$ is continuous, $\mathbb{D}(n;\mathbb R)$ cannot be dense.

Gabriel Romon
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  • For (c) $A=\begin{pmatrix}0&0\0&n+1 \end{pmatrix}$ i judged because, for $\forall n\in \mathbb{N}$ $\exists$ a matrix like this whose norm greater than n. Am I correct? –  Nov 02 '17 at 11:03
  • (d) Using the continuity of $f$, how can I say $\mathbb{D}(n,\mathbb{R})$ is not dense? –  Nov 02 '17 at 11:15
  • @ManeeshNarayanan reason by contradiction – Gabriel Romon Nov 02 '17 at 11:16
  • We want to prove " $f$ is continuous $\implies$ $\mathbb{D}(n,\mathbb{R})$ is not dense Suppose $\mathbb{D}(n,\mathbb{R})$ is dense $\implies$ $A\in M(n,\mathbb{R})$, $\exists$ a sequence in $\mathbb{D}(n,\mathbb{R})$ , which converges to $A$. This much I can draw. How to proceed further? –  Nov 02 '17 at 11:24
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    @ManeeshNarayanan first thing, $n=2$. Then choose some $A$ with no real eigenvalue. Hence $f(A)<0$. For a sequence $B_n$ that converges to $A$, $f(B_n)\geq 0$ for all $n$, and by continuity of $f$, $0 \leq f(B_n)\to f(A)<0$, a contradiction. – Gabriel Romon Nov 02 '17 at 11:27
  • ok. I got it. Thank you very much. –  Nov 02 '17 at 11:27
  • How to prove for $n$? –  Nov 02 '17 at 11:55