# 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?

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...

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.

• 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? – Connor Robetorye Jan 16 at 2:51
• 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". – user299843 Jan 16 at 16:13