Are all diagonalizaitions necessarily eigendecompositions? If a matrix $A \in \mathbb{F^{NxN}}$ can be diagonalized, i.e. factored into the form:
$A = V \Lambda V^{-1}$, where V is a basis for $\mathbb{F^N}$ and $\Lambda$ is a diagonal matrix, do $V$ and $\Lambda$ necessarily contain the eigenvectors and eigenvalues of $A$ respectively, or could a diagonalization be in terms of other vectors/diagonal values?
Is the answer to this question different between asymmetric square matrices on the one hand and symmetric/Hermitian or normal matrices which decompose as $Q\Lambda Q'$ for unitary $Q$ on the other hand? And if so, or if all diagonalizations are necessarily in terms of eigenvectors/values, why?
 A: Rewrite
$$
A = V \Lambda V^{-1}
$$
as
$$
AV = V \Lambda
$$
and let $v_1$ be the first column of $V$.
Now look at the first column of $AV$ --- it's just $Av_1$.
What about the first column of $V \Lambda$? It's just $\lambda_1 v_1$.
The same reasoning applies to all other columns, hence each column of $V$ is an eigenvector of $A$.
I leave you to think about the complex case on your own...
A: Since $A=V\Lambda V^{-1}$ the eigenvalues of $A$ and $\Lambda$ are the same, since $\Lambda$ is diagonal, we can read the eigenvalues immediately off from $\Lambda$. Writing the equation as $AV=V\Lambda$, then if $\Lambda=\text{diag}(\lambda_1,\ldots,\lambda_n)$ and the column vectors of $V$ are $v_1,\ldots,v_n$, we see that $Av_i=\lambda_i v_i$ holds, so we can read off from $V$ an eigenvector $v_i$ corresponding to the eigenvalue $\lambda_i$ for every $i$.
A diagonalisation in terms of different $V$ and $\Lambda$ is possible. For instance, we could permute the entries of $\Lambda$. Also, if $v$ is an eigenvector, so is $\lambda v$ for a non-zero scalar $\lambda$. Therefore, we could multiply every column of $V$ with a different non-zero scalar and still get a diagonalisation.
