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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?

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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...

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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.

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  • $\begingroup$ If you were to permute the elements of lambda you would also have permute the corresponding columns of V to preserve equality, right? And since scaled columns of V are still eigenvectors, what I'm understanding is that any diagonalization is necessarily an eigendecomp, but not all eigendecomps are unique, correct? $\endgroup$ – Connor Robetorye Jan 16 at 2:51
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    $\begingroup$ Indeed, you would have to permute the corresponding columns as well. Exactly, no eigendecomposition is unique (there is always some freedom left). Consider for instance the eigendecomposition of $I_n$: we have $I_n=V I_n V^{-1}$ for all $V\in\text{GL}_n$, lots of freedom! In general only the diagonal matrix is sort of "unique". $\endgroup$ – user299843 Jan 16 at 16:13

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