Can non-normal matrices with double eigenvalues never be diagonalized?

Question

Is there a matrix $A$ with $A^TA≠ AA^T$ (non-normality) and double eigenvalue that is still diagonalizable?

If $A^TA \neq AA^T$ and $λ_1 =λ_2 = λ$ (double eigenvalue)

$\stackrel{?}{⇒}$

not exists $V$, $Ω$ with $A = V Ω V^{-1}$ ?

As one example consider $\begin{bmatrix} 1 & 1\\ 0 & a \end{bmatrix}$ which is obviously non-normal. When I change the matrix to $\begin{bmatrix} 1 & 1\\ 0 & 1 \end{bmatrix}$ in order to get a double eigenvalue, it turns out to become non-diagonalizable. Is this always the case?

Answer 1

In dimensions $3$ and higher, it is very easy to construct an example like you want. Let us start with a diagonal matrix $$ D=\begin{bmatrix}1&0&0\\0&1&0\\0&0&0\end{bmatrix}. $$ If we now conjugate $D$ with a non-symmetric invertible matrix, we will likely get an example like the one you want. Say we take $$ S=\begin{bmatrix}1&1&1\\0&1&1\\0&0&1\end{bmatrix}. $$ Then $$ A=SDS^{-1}=\begin{bmatrix}1&1&1\\0&1&1\\0&0&1\end{bmatrix}\begin{bmatrix}1&0&0\\0&1&0\\0&0&0\end{bmatrix}\begin{bmatrix}1&-1&0\\0&1&-1\\0&0&1\end{bmatrix}=\begin{bmatrix}1&0&-1\\0&1&-1\\0&0&0 \end{bmatrix}. $$ Then $$ A^TA=\begin{bmatrix}1&0&-1\\0&1&-1\\-1&-1&2\end{bmatrix},\ \ AA^T=\begin{bmatrix}2&1&0\\1&2&0\\0&0&0\end{bmatrix}. $$

Answer 2

Let $A=\begin{bmatrix} 0 & 0 & 1 \\ 0 & 0 & 1 \\ 0 & 0 & 1 \end{bmatrix}$. It is easy to check that $A$ is not normal, but using $V=\begin{bmatrix} 1 & 1 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{bmatrix}$, we can check that $V^{-1} A V= \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 1 \end{bmatrix}$.

Answer 3

Maybe the simplest example (in the sense of, fewest non-zero entries) would be $$A=\pmatrix{0&0&0\cr0&0&1\cr0&0&1\cr}$$