If $K$ is an invertible matrix so $K+K^{-1}$ is diagonalizable, is $K$ necessarily diagonalizable?

Question

Let $K$ be an invertible $n\times n$ matrix so $K+K^{-1}$ is diagonalizable. Is $K$ necessarily diagonalizable?

Can't come up with a counter example, but also can't prove it... Here's my attempt so far:

At first i proved that if $K$ is invertible and diagonalizable so that $K=P^{-1}AP$ ($A$ being a diagonal matrix) then $K+K^{-1}$ is also diagonalizable, and $K+K^{-1}=P^{-1}\left(A+A^{-1}\right)P$. Then I tried using the fact that there's a diagonal matrix $B$ so that:

$B=P^{-1}\left(K+K^{-1}\right)P=P^{-1}KP+P^{-1}K^{-1}P$

So if we name them $P^{-1}KP=C\:\:\:\&\:\:\:\:P^{-1}K^{-1}P=D$ we get something like:

$B=C+D=diag\left(c_{11}+d_{11},\:c_{22}+d_{22},\:...,\:c_{nn}+d_{nn}\right)$

And since $CD=I_n$ I thought it would be useful to display it this way:

$C\left(diag\left(...\right)-C\right)=C\:diag\left(...\right)-C^2=D\:diag\left(...\right)-D^2=I_n$

And now I'm completely stuck.

Answer 1

\begin{pmatrix} 1 & 1 \\ 0 & 1\\ \end{pmatrix} seems to work as a counter example. It is bviously not diagonalisble (Jordan normal form) but it's inverse is \begin{pmatrix} 1 & -1 \\ 0 & 1\\ \end{pmatrix} so the sum of those two matrices is \begin{pmatrix} 2 & 0 \\ 0 & 2\\ \end{pmatrix}.

Answer 2

Here is a counter example

$$ K=\begin{bmatrix} 1&0\\1&1 \end{bmatrix} $$ which is not diagonalizable but $$ K+K^{-1}=\begin{bmatrix} 1&0\\1&1 \end{bmatrix}+\begin{bmatrix} 1&0\\-1&1 \end{bmatrix}=\begin{bmatrix} 2&0\\0&2 \end{bmatrix} $$ which is diagonalizable.

Answer 3

$A\in Gl_2(\Bbb{R}) $ where

$A=\begin{pmatrix} 0 & -1 \\1&0\end{pmatrix}$

$A^{-1}=\begin{pmatrix} 0 & 1 \\-1&0\end{pmatrix}$

$A+A^{-1}=\begin{pmatrix} 0 & 0 \\0&0\end{pmatrix}$ diagonalizable but $A$ is not diagonalizable over $\Bbb{R}$