Let $B\in GL_n (\mathbb{R})$ such that :

Question

$B-B^{-1}$ is diagonalizable. Show that $B$ is diagonalizable.

Let $P(X)=\prod \limits_{i=1}^{r}(X-\lambda_i)$ the minimal polynomial of $B-B^{-1}$.

Moreover : $\prod \limits_{i=1}^{r}(B-B^{-1}-\lambda_i)=0$

Then maybe, we can multiply by $B$, $r$ times. So we will obtain :

$\prod \limits_{i=1}^{r}(B^2-\lambda_iB-I_n)=0$

It gives a polynomial which $B$ is the indeterminate and which vanished in $B$.

But how to continue ?

Thanks in advance !

Answer 1

Recall that a matrix is diagonalisable over a field $F$ if and only if its minimal polynomial can be factored into distinct linear factors with coefficients in $F$.

So, as $A=B-B^{-1}$ is diagonalisable over $\mathbb R$, its minimal polynomial can be written as $f(x)=\prod_{i=1}^r(x-\lambda_i)$, where $\lambda_1,\lambda_2,\ldots,\lambda_r$ are some $r$ distinct real numbers. As you said, from $f(A)B=0$, we obtain $p(B)=0$ where $p(x)=\prod_{i=1}^r(x^2-\lambda_ix-1)$.

We want to show that $B$ is diagonalisable over $\mathbb R$, i.e. we want to show that its minimal polynomial --- denoted by $g(x)$ in the sequel --- can be factored into distinct linear factors with real coefficients. However, by definition, $g(x)$ must divide $p(x)$. Therefore it suffices to prove that $p(x)$ can be factored into distinct linear factors.

In other words, it suffices to prove that all roots of $\prod_{i=1}^r(x^2-\lambda_ix-1)=0$ are real and distinct. Note that what one needs to show is not only that the roots of $x^2-\lambda_ix+1=0$ are real and distinct for each $i$, but also that the roots of $x^2-\lambda_ix-1=0$ are different from those of $x^2-\lambda_jx-1=0$ when $i\ne j$.

Anyway, this boils down to a calculus exercise that is not hard to do. For starters, note that both $\lambda\mapsto \lambda-\sqrt{\lambda^2+4}$ and $\lambda\mapsto \lambda-\sqrt{\lambda^2+4}$ are strictly increasing functions on $\mathbb R$ and $\lambda+\sqrt{\lambda^2+4}>0>\mu-\sqrt{\mu^2+4}$ for all real numbers $\lambda$ and $\mu$.

Answer 2

Let $spectrum(B)=(\lambda_i)_i$ where $\lambda_i\not= 0$. Then $spectrum(B-B^{-1})=(\lambda_i-1/\lambda_i)_i$.

i) The $(\lambda_i)$ are real. Indeed, if $\lambda-1/\lambda=a\in\mathbb{R}$, then $\lambda=\dfrac{a\pm \sqrt{a^2+4}}{2}\in\mathbb{R}$.

ii) Assume that $B$ is not diagonalizable. Then the Jordan decomposition of $B$ contains a factor of the type $J=\lambda I+N$ where $\lambda\not= 0$ and $N$ is non-zero nilpotent. It remains to show that $J-J^{-1}$ is not diagonalizable.

Since $J^{-1}=1/\lambda I-1/\lambda^2N+\cdots$, we deduce that $J-J^{-1}=(\lambda-1/\lambda)I+(1+1/\lambda^2)N+\cdots$, that is not diagonalizable, because $1+1/\lambda^2\not= 0$.

EDIT. We can generalize as follows: Let $f:U\subset \mathbb{C}\mapsto \mathbb{C}$ be an analytic function s.t.

i) $f(z)\in\mathbb{R}\implies z\in\mathbb{R}$

ii) If $z\in\mathbb{R}$, then $f'(z)\not= 0$.

$\textbf{Proposition}$. Let $B\in M_n(\mathbb{R})$ s.t. $spectrum(B)\subset U$; if $f(B)$ is diagonalizable over $\mathbb{R}$, then $B$ too.