# Simultaneous diagonalization

I'm pretty sure that the following (if true) is a standard result in linear algebra but unfortunately I could not find it anywhere and even worse I'm too dumb to prove it: Let $k$ be a field, let $V$ be a finite-dimensional $k$-vector space and let $S \subseteq \mathrm{End}_k(V)$ be a subset of pairwise commuting (i.e. $\lbrack S, S \rbrack = 0$) endomorphisms. Then the following holds:

1. If all $f \in S$ are diagonalizable, then there exist maps $\chi_i:S \rightarrow k$, $i=1,\ldots,r$, such that $V = \bigoplus_{i=1}^r E_{\chi_i}(S)$, where $E_\chi(S) := \lbrace v \in V \mid fv = \chi(f)v \ \forall \ f \in S \rbrace$.

2. The maps $\chi_i$ in 1 are unique.

3. 1 is equivalent to the existence of a basis $\mathcal{B}$ of $V$ such that for each $f \in S$ the matrix $M_{\mathcal{B}}(f)$ of $f$ with respect to $\mathcal{B}$ is diagonal. (I believe that this might not be true)

4. If all $f \in S$ are trigonalizable, then there exists a basis $\mathcal{B}$ of $V$ such that for each $f \in S$ the matrix $M_{\mathcal{B}}(f)$ of $f$ with respect to $\mathcal{B}$ is upper triangular and for each diagonalizable $f \in S$ the matrix $M_{\mathcal{B}}(f)$ is diagonal.

I know that a set of commuting diagonalizable endomorphisms can be simultaneously diagonalized in the sense of 3 but I don't know how to prove 1 (my problem is the "glueing" of the $\chi$-maps when I try to prove this by induction on $\mathrm{dim}V$). Also, I know that the first part of 4, the simultaneous trigonalization, holds but I don't know how to show that there exists a basis which then also diagonalizes all diagonalizable endomorphisms. This should follow from 1, I think.

Perhaps, because all this is probably standard stuff, I should mention that this is not a homework problem :)

One additional question: Suppose that $k$ is algebraically closed and that $G$ is an affine commutative algebraic group over $k$ which coincides with its semisimple part, embedded as a closed subgroup in some $GL(V)$. Are the maps $\chi_i:G \rightarrow \mathbb{G}_{m}$ morphisms of algebraic groups?

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All of these are true. First note that the space of endomorphisms of $V$ is finite-dimensional, so even an infinite $S$ can just be replaced by finitely many matrices that have the same span (it's really more elegant to think about the span of $S$ as a Lie algebra, rather than $S$ itself). You actually may want to look at some discussion of abelian Lie algebras, since really your question is about the natural structure theorem for semi-simple representations of abelian Lie algebras (if you think in this language question 4) is obvious from the first 3, since any representation has a flag whose successive quotients are semi-simple).

The important point for proving 1) is that if A and B commute and are both diagonalizable, you should analyze the action of B on the eigenspaces of A. The spaces $E_\chi$ above are the eigenspaces of B's action on each eigenspace of A (and if there were a third matrix, you would take the eigenspaces of C acting on the eigenspaces of B in the eigenspaces of A, etc.). 2) is clear, and for 3) just pick any basis of these iterated eigenspaces.

For 3) => 1), you have associated to each basis vector a $\chi$, given by looking at how the elements of $S$ act on it. $E_\chi$ is just the span of all vectors associated to the particular map $\chi$.

For 4), there is a similar argument, but you have to use a flag (the $i$-th subspace being things killed by $(A-\lambda I)^i$ for some scalar $\lambda$) rather than a subspace decomposition. Still, matrices commuting means that B will preserve this flag, so just as one refined the eigenspaces, one can refine this flag.

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I'm glad that all is true. However, I'm still not able to prove this. For 1: Do you somehow need that $S$ is finite? Because my $S$ might be infinite. For 3: This proves the obvious direction 1=>3, but what about 3=>1? I don't know how to get the chi-maps from a basis consisting of simultaneous eigenvectors. –  user717 Dec 22 '09 at 16:27
3. Is there any restriction on your \chi at all? If not, it's obvious - say your basis of eigenvectors is {e_1,...,e_n}. Then E_i = span{e_i} would work. I don't know how you want to express your chi-maps, but I think \chi_i(f) = the eigenvalue of e_i corresponding to f would be okay. –  Ho Chung Siu Dec 22 '09 at 16:56
1. S could be infinite as long as V is of finite dimension. Just like what Ben has said, you can just look at induced action on eigenspaces. For example, take f in S, and write V = \sum E_i a nontrivial eigenspace decomposition. (assuming that f is not a scalar for now) - If for the rest of the endomorphisms in S, each E_i is an eigenspace, we are done. - If not, WLOG say for g in S, g|E_i is not scalar. Then you break E_1 further into dierct sum of eigenspace of g. This way the number of "components" increases. This number cannot exceed dim V, so this process must stop somehow. –  Ho Chung Siu Dec 22 '09 at 16:56
@Ho Chung Siu: But E_i = span{e_i} is not necessarily of the form E_chi because these do not necessarily have to be one-dimensional!? S={f} with a two-fold eigenvalue would be such a case I think. –  user717 Dec 22 '09 at 17:12
@Arminius So you want the \chi_i in 1 to be all distinct? In that case, look at the \chi_i associated to each e_i. Put all the e_i with the same \chi function together and span these e_i. Then that should work. –  Ho Chung Siu Dec 22 '09 at 17:42

And the answer to the additional question (which Ben skipped) is also positive. Indeed, there are two parts here:

1. $\chi_i$ is a group morphism (this is obvious for any subgroup $G\subset GL(V)$

2. $\chi_i$ is algebraic (this is obvious because matrix elements of matrices in $GL(V)$ in any basis are algebraic functions for tautological reasons).

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Why exactly are the chi_i algebraic?? –  user717 Dec 22 '09 at 17:18
Choose a basis such that the first vector is in $E_{\chi_i}$, then $\chi_i$ is the 1-1 matrix element. –  t3suji Dec 22 '09 at 17:32
Oh dear! Thanks. –  user717 Dec 22 '09 at 17:41

Hi Arminius, For part (1) I believe you're asking that if two diagonalizable matrices commute, then they have the same eigenspaces (aka they are simultaneously diagonalizable). I believe this is true if you make all the eigenvalues distinct:

Let $D$ be diagonal with distinct entries, then if it commutes with $A$:

$AD$ = $DA$

on the LHS of the equation, all the columns of $A$ are scaled by different factors. On the RHS the columns are scaled. The only way to reconcile this is if $A$ is diagonal too.

Hope this helps.

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