The product diagonalisable matrices of determinant 1 is not necessarily diagonalisable

Question

I need to confirm, for a proof, a general result of linear algebra:

Let $g, h \in SL_n(\mathbb{C})$ two diagonalisable (semisimple) matrices. The product of these matrices is not necessarily diagonalisable.

I've tried to write $g = xdx^{-1}$ and $h= yd'y^{-1}$ where $x,y \in GL_n(\mathbb{C})$ and $d$ and $d'$ two diagonal matrices.

We have $gh = xdx^{-1}yd'y^{-1}$. This is clearly not with the form of product of an invertible matrix and diagonal matrix and the inverse of the first matrix.

But I wonder if there is better argument for this. I tried to think about an example in $SL_n(\mathbb{C})$ but I didn't find.

Thanks in advance for your enlightements.

K. Y.

Answer 1

If $a \ne a^{-1}$, then $$ \begin{bmatrix} a & a \\ 0 & a^{-1}\end{bmatrix} $$ is diagonalizable but $$ \begin{bmatrix} a^{-1} & 0 \\ 0 & a\end{bmatrix} \begin{bmatrix} a & a \\ 0 & a^{-1}\end{bmatrix} = \begin{bmatrix} 1 & 1 \\ 0 & 1\end{bmatrix} $$ is not.

Answer 2

Your argument is not valid; as suggested in the comments, just because $gh$ might have a form that is not diagonal, it does not imply that $gh$ is not diagonalizable.

An example of two matrices with determinant -1 multiplying to give a non-diagonal matrix is given in: Is the product of any two invertible diagonalizable matrices diagonalizable?

You can multiply both matrices by -1 and that should give you your counterexamples.

Answer 3

$$ \left( \begin{array}{cc} 2 & \frac{1}{4} \\ 0 & \frac{1}{2} \end{array} \right) \left( \begin{array}{cc} \frac{1}{2} & \frac{1}{4} \\ 0 & 2 \end{array} \right) = \left( \begin{array}{cc} 1 & 1 \\ 0 & 1 \end{array} \right) $$