Let $V$ be a vector space of finite dimension and let $T,S$ linear diagonalizable transformations from $V$ to itself. I need to prove that if $TS=ST$ every eigenspace $V_\lambda$ of $S$ is $T$-invariant and the restriction of $T$ to $V_\lambda$ ($T:{V_{\lambda }}\rightarrow V_{\lambda }$) is diagonalizable. In addition, I need to show that there's a base $B$ of $V$ such that $[S]_{B}^{B}$, $[T]_{B}^{B}$ are diagonalizable if and only if $TS=ST$.

Ok, so first let $v\in V_\lambda$. From $TS=ST$ we get that $\lambda T(v)= S(T(v))$ so $T(v)$ is eigenvector of $S$ and we get what we want. I want to use that in order to get the following claim, I just don't know how.

One direction of the "iff" is obvious, the other one is more tricky to me.

I'd love your help with this.


  • $\begingroup$ $0$ is diagonalizable. $\endgroup$ – Qiaochu Yuan Aug 8 '11 at 14:59

This answer is basically the same as Paul Garrett's. --- First I'll state the question as follows.

Let $V$ be a finite dimensional vector space over a field $K$, and let $S$ and $T$ be diagonalizable endomorphisms of $V$. We say that $S$ and $T$ are simultaneously diagonalizable if (and only if) there is a basis of $V$ which diagonalizes both. The theorem is

$S$ and $T$ are simultaneously diagonalizable if and only if they commute.

If $S$ and $T$ are simultaneously diagonalizable, they clearly commute. For the converse, I'll just refer to Theorem 5.1 of The minimal polynomial and some applications by Keith Conrad.

EDIT. The key statement to prove the above theorem is Theorem 4.11 of Keith Conrad's text, which says:

Let $A: V \to V$ be a linear operator. Then $A$ is diagonalizable if and only if its minimal polynomial in $F[T]$ splits in $F[T]$ and has distinct roots.

[$F$ is the ground field, $T$ is an indeterminate, and $V$ is finite dimensional.]

The key point to prove Theorem 4.11 is to check the equality $$V=E_{\lambda_1}+···+E_{\lambda_r},$$ where the $\lambda_i$ are the distinct eigenvalues and the $E_{\lambda_i}$ are the corresponding eigenspaces. One can prove this by using Lagrange's interpolation formula: put $$f:=\sum_{i=1}^r\ \prod_{j\not=i}\ \frac{T-\lambda_j}{\lambda_i-\lambda_j}\ \in F[T]$$ and observe that $f(A)$ is the identity of $V$.

  • 1
    $\begingroup$ Oh, I can only award you in 23 hours. we'll wait :) $\endgroup$ – user6163 Aug 16 '11 at 19:35
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    $\begingroup$ @Nir - Dear Nir, you’re very generous. I’m deeply moved by your decision. I’m certainly doing my best, but I’m sure that’s what everybody is doing here. Let’s try to concentrate on mathematics and human relationships, and let’s forget points and reputation and comparisons. Also, it was a pleasure to work on your so interesting questions... But there are only two words I can say: thank you! $\endgroup$ – Pierre-Yves Gaillard Aug 16 '11 at 20:27
  • $\begingroup$ Thanks for providing the document. It is enlightening. $\endgroup$ – awllower Nov 30 '11 at 8:37
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    $\begingroup$ You've listed a document by Keith Conrad. I believe this one by the same author also has a very simple proof for the question being asked without reference to a minimal polynomial. I found it helpful since I didn't know what a minimal polynomial is: math.uconn.edu/~kconrad/blurbs/linmultialg/simulcomm.pdf $\endgroup$ – Sherif F. Sep 24 '14 at 17:26

You've proven (from $ST=TS$) that the $\lambda$-eigenspace $V_\lambda$ of $T$ is $S$-stable. The diagonalizability of $S$ on the whole space is equivalent to its minimal polynomial having no repeated factors. Its minimal poly on $V_\lambda$ divides that on the whole space, so is still repeated-factor-free, so $S$ is diagonalizable on that subspace. This gives an induction to prove the existence of a simultaneous basis of eigenvectors. Note that it need not be the case that every eigenvector of $T$ is an eigenvector of $S$, because eigenspaces can be greater-than-one-dimensional.

Edit: thanks @Arturo M., yes, over a not-necessarily algebraically closed field, one must say that "diagonalizable" is eq't to no repeated factor _and_splits into linear factors.

Edit.2: $V_\lambda$ being "S-stable" means that $SV_\lambda\subset V_\lambda$, that is, $Sv\in V_\lambda$ for all $v\in V_\lambda$.

  • $\begingroup$ I think you want to add "splits" to the condition on the minimal polynomial. The rotation by $\pi/2$ in $\mathbb{R}^2$ has minimal polynomial $x^2+1$ which has no repeated factors, but is not diagonalizable. $\endgroup$ – Arturo Magidin Aug 8 '11 at 15:52
  • $\begingroup$ What is $S$-stable? $\endgroup$ – user6163 Aug 8 '11 at 16:38
  • $\begingroup$ @Nir: "$S$-stable" is the same as "$S$-invariant": the restriction of $S$ induces an operator/endomorphism. $\endgroup$ – Arturo Magidin Aug 9 '11 at 18:25

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