We are given the following situation:

Suppose $T_1$ and $T_2$ are linear operators on a finite dimensional vector space $V$ satisfying $T_1T_2 = T_2T_1$, and there are bases $B_1$ and $B_2$ such that $[T_1]_{B_1}$ and $[T_2]_{B_2}$ are both diagonal.

I want to show that if we have an Eigenvector of one of these linear transformations then it is an Eigenvector of the other. Suppose that $T_1x=\lambda x$

$$ \begin{align*} T_1T_2 x &= T_2T_1 x \\ T_1T_2 x &= T_2(\lambda x) \\ T_1T_2 x &= \lambda T_2x \end{align*} $$

This implies that $T_2x$ is an eigenvector of $T_1$, and that $T_2x \in E_\lambda$ where $E_\lambda$ is the $\lambda$-Eigenspace of $T_1$. But if this is true then that means that $x$ must be an Eigenvector of $T_2$. A symmetrical argument shows that if $x$ is an eigenvector of $T_2$ then we have that $x$ must also be an eigenvector of $T_1$.

I am unsure of my last step here though can I conclude that $x$ is an Eigenvector of $T_2$ if $T_2x$ is in the $\lambda$-Eigenspace, it seems pretty iffy...


1 Answer 1


Start with $T_1x=\lambda x$ and consider what is $T_1y$ with $y=T_2x$. Then easily $$ T_1y=T_1T_2x=T_2T_1x=T_2\lambda x =\lambda T_2 x = \lambda y $$ since $T_1$ and $T_2$ commute, so $y=T_2x$ is an eigenvector of $T_1$ with the same eigenvalue as $x$.

Suppose the eigenvalue $\lambda$ occurs only once. Then it must follow that $y=T_2x$ is a multiple of $x$ since $x$ is, up to any multiplicative factor, eigenvector of $T_1$ with eigenvalue $\lambda$. If $y$ is a multiple of $x$ it means $y=T_2x = \lambda_2x$, meaning that $x$ is an eigenvector of $T_2$ with eigenvalue $\lambda_2$.

If the eigenvalue $\lambda$ is repeated, the argument becomes more complicated as you can only conclude that $T_2x$ is a linear combination of eigenvectors of $T_1$, ie. $T_2x=a_1y_1+a_2y_2$, with $T_1y_1=\lambda y_1$ and $T_1y_2=\lambda y_2$. At this point the proof that you can find simultaneous eigenvectors of $T_1$ and $T_2$ proceeds by recursion.


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