TITLE: Can non-normal matrices with double eigenvalues never be diagonalized?
QUESTION [7 upvotes]: 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?

REPLY [4 votes]: Maybe the simplest example (in the sense of, fewest non-zero entries) would be $$A=\pmatrix{0&0&0\cr0&0&1\cr0&0&1\cr}$$