Can non-normal matrices with double eigenvalues never be diagonalized? 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?
 A: 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
$$
A^TA=\begin{bmatrix}1&0&-1\\0&1&-1\\-1&-1&2\end{bmatrix},\ \ AA^T=\begin{bmatrix}2&1&0\\1&2&0\\0&0&0\end{bmatrix}.
$$
A: Maybe the simplest example (in the sense of, fewest non-zero entries) would be $$A=\pmatrix{0&0&0\cr0&0&1\cr0&0&1\cr}$$
A: 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 $V^{-1} A V= \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 1 \end{bmatrix}$.
