4

I know it's been answered before (at least to the case with $n$ different eigenvalues) but I didn't find a proof for the general case, and I would like some help with this question.

We are given linear transforms $S,T: V\to V$ where $V$ is some vector space.

We are given that $S$ and $T$ commute, $ST=TS$, and that they are diagonalizable:

$T=PD_1P^{-1}$ and $S=KD_2K^{-1}$, where $D_1, D_2$ are diagonal and $K,P$ are invertible.

We are asked to show that $S$ and $T$ have a common eigenspace.

My solution

Maybe I understood the question wrong, but what I tried to do is show that if $v$ is an eigenvector of $S$ then it is also an eigenvector of $T$.

let $Sv=\lambda v$.

$STv=TSv=T\lambda v=\lambda Tv$ which implies that $Tv$ is an eigenvector of $S$ with eigenvalue $\lambda$.

Why does that mean that $v$ is an eigenvalue of $T$?

Another possible way to solve this question is write:

$PD_1P^{-1}KD_2K^{-1} = KD_2K^{-1}PD_1P^{-1}$ and get that $P=K$ but I don't know how to do that either.

glS
  • 6,818
Oria Gruber
  • 12,739

4 Answers4

9

The result which's known for two (or more) diagonalizable matrices that commute is that are simultaneous diagonalizable, that's they are diagonalizable in the same basis. Let's prove it by induction on the dimension $\dim E$ (the result is trivial if $\dim E=1$):

If $S$ or $T$ is an homothetie then the result is obvious. Now assume that neither $S$ nor $T$ is an homothetie and since $S$ is diagonalizable then $$E=\bigoplus_{\lambda\in\mathrm{sp}(S)}E_\lambda(S)$$ where $E_\lambda(S)$ is the eigenspace of $S$ associated to the eigenvalue $\lambda$. Since $S$ isn't an homothetie then $$\forall\lambda\in \mathrm{sp}(S)\;\;\dim E_\lambda(S)\le\dim E-1$$ and since $ST=TS$ then $ E_\lambda(S)$ is invariant by $T$. Let $T'=T_{| E_\lambda(S)}$ the restriction of $T$ to $ E_\lambda(S)$ so by hypothesis $S$ and $T'$ are simultaneous diagonalizable on $ E_\lambda(S)$ that's there's a basis $B_\lambda$ of $ E_\lambda(S)$ in which $T'$ is diagonal. In the basis $B=\cup B_\lambda$ the two matrices $S$ and $T$ are diagonal.

  • could you elaborate why for any lambda: $dimE_{\lambda}(S) \leq dimE-1$? What if there is only 1 eigenvalue? In that case since the transformation is diagonlizable, we will get $dimE_{\lambda}(S)=dimE$ I should note that I don't know what a hmothetie is. I've never heard that term before and so I may be speaking nonsense :P But i would like a clarification – Oria Gruber Dec 25 '13 at 22:07
  • The transformation $\alpha\mathrm{id}$ is a homothetie and we have $$E_\lambda(S)=E\iff S=\lambda\mathrm{id}$$ –  Dec 25 '13 at 22:13
7

If $S$ and $T$ commute and both are diagonalizable then there is a common transform that will simultaneously diagonalize both.

First find a transform $M$ so that $M^{-1}SM$ is diagonal. Suppose that the eigenvalues of $S$ are sorted so repeated eigenvalues of $S$ are grouped together.

Let us suppose that $\lambda_1$ has multiplicity $m$. Since the eigenspace of $S$ is invariant under $T$ we should have $$\begin{align} M^{-1} S M &= \begin{pmatrix}\lambda I_m & 0 \\ 0 & \hat{S}_{22}\end{pmatrix} \\ M^{-1} T M &= \begin{pmatrix}\hat{T}_{11} & 0 \\ 0 & \hat{T}_{22}\end{pmatrix} \end{align}$$ where $I_m$ is $m \times m$ identity matrix. Let $P$ diagonalize $\hat{T}_{11}$ i.e. $$ P^{-1} \hat{T}_{11} P = D $$ where $D$ is diagonal.

Now let $$ R= M \begin{pmatrix} P & 0 \\0 & I\end{pmatrix} $$ Then $$ R^{-1} S R = \begin{pmatrix} P^{-1} & 0 \\0 & I\end{pmatrix} \begin{pmatrix}\lambda I_m & 0 \\ 0 & \hat{S}_{22}\end{pmatrix} \begin{pmatrix} P & 0 \\0 & I\end{pmatrix} = \begin{pmatrix}\lambda I_m & 0 \\ 0 & \hat{S}_{22}\end{pmatrix} $$ $$ R^{-1} T R = \begin{pmatrix} P^{-1} & 0 \\0 & I\end{pmatrix} \begin{pmatrix}\hat{T}_{11} & 0 \\ 0 & \hat{T}_{22}\end{pmatrix} \begin{pmatrix} P & 0 \\0 & I\end{pmatrix} = \begin{pmatrix}D & 0 \\ 0 & \hat{T}_{22}\end{pmatrix} $$ Thus the $m\times m$ block is now simultaneously diagonalized.

Proceed as before for other eigenvalues of $S$.

user44197
  • 9,730
0

The result is not true. The linear transformations of a space of dimension$~3$ with, with respect to a fixed basis, have diagonal matrices with diagonal entries $(0,0,1)$ respectively $(0,1,1)$ satisfy all hypotheses, but they do not have an eigenspace in common.

Marc van Leeuwen
  • 115,048
0

You are almost there...

You have shown that the eigenspace of $S$ corresponding to $\lambda$ is invariant under $T$. If $\lambda$ is an eigenvalue of multiplicity 1, then $Tv$ is a multiple of $v$ and hence is an eigenvalue of $T$.

If this is not the case, all you can say is the invariance of the eigenspaces. Take for example $S=I$ the identity matrix. Every vector is an eigenvector of $S$ but this is not true for a general $T$.

The result is not true in general

user44197
  • 9,730
  • That is exactly my problem. We weren't given that there are $n$ eigenvalues and thus we can't assume or know anything about the multiplicity of $\lambda$. I agree that if $V_{\lambda}$ is an eigenspace of $S$ then $T(V_{\lambda})$ is also an eigenspace of $S$. But why does that mean that $V_{\lambda}$ is an eigenspace of $T$? wikipedia clearly says ": A set of diagonalizable matrices commutes if and only if the set is simultaneously diagonalisable" so the statement is correct. – Oria Gruber Dec 25 '13 at 21:38
  • The example you gave is false because: We were given that $S,T$ are diagonlizable. If $S=I$ then every vector is an eigenvector. but we know that $T$ is also diagonlizable, so we found a shared basis of eigenvectors, the eigenvectors of $T$. – Oria Gruber Dec 25 '13 at 21:42
  • 1
    Let me amplify what I wrote: While it is true that there exists a common transformation will simultaneously diagonalize both, it is not true that any transformation that diagonalizes one will also diagonalize the other. I will provide a more complete answer soon. – user44197 Dec 25 '13 at 21:59