Given A a square matrix and its SVD decomposition A=UEV. It is possible to prove that EV is always diagonalizable?

Question

I am trying to prove or ensure that this is not true:

Given A a square matrix and its SVD decomposition A=UEV. Is it possible to prove that EV is always diagonalizable?

Does someone know if this statement is valid?

Answer 1

HINT:

Your statement is not true in general. In fact, every invertible matrix is conjugate for a matrix of the form $D \cdot V$. Indeed, every invertible matrix can be written as $$A = P \cdot W$$ where $P$ is positive definite and $W$ is unitary ( orthogonal in the real case) -- see polar decomposition. Now, we can conjugate $P$ by a unitary(orthogonal) to a diagonal matrix with positive elements, so $R P R^{-1} = D$. Now we get $$R A R^{-1} = R P R^{-1} \cdot R W R^{-1} = D \cdot V$$

For a concrete example, look at $2\times 2$ matrices. Search for a product of the form $$\left( \begin{matrix} p & 0 \\ 0 & q\end{matrix} \right ) \cdot \left( \begin{matrix} a & -b \\ b & a \end{matrix}\right )$$ that has equal eigenvalues. Check that $$\left( \begin{matrix} 2 & 0 \\ 0 & 1/2\end{matrix} \right ) \cdot \left( \begin{matrix} 4/5 & -3/5 \\ 3/5 & 4/5 \end{matrix}\right )$$ has equal eigenvalues $1$, $1$. Or see a general counterexample here.

$\bf{Added:}$ The above example of a unipotent matrix can be written as: $$\left( \begin{matrix} \frac{1+\tan u}{1-\tan u} & 0 \\ 0 & \frac{1-\tan u}{1+ \tan u} \end{matrix} \right ) \cdot \left( \begin{matrix} \cos 2u & -\sin 2u \\ \sin 2 u & \cos 2 u \end{matrix}\right )$$