Let $V$, an $n$-dimensional vector space and let $T, S:V\to V$, two diagonalizable linear operators. Show that if $TS=ST$ then every $V_\lambda$ of $S$ is $T$-invariant and the restriction, $T|_{V_\lambda}$ is diagonalizable.

So first it easy to show that $V_\lambda$ (an eigensapce of $S$) is $T$-invariant. Let $v\in V_\lambda$. Then

$$S(T(v)) = T(S(v)) = T(\lambda v) = \lambda T(v)$$

So indeed, $T(v)$ is an eigenvector of $V_\lambda$.

Now, I need to show that $T|_{V_\lambda}$ is diagonalizable.

Since $ST=TS$ and $S,T$ are diagonalizable then we know that there's a basis $B$ such that $$[T]_B = \text{Diag}(\lambda_1, \ldots ,\lambda_n) \\ [S]_B = \text{Diag}(\mu_1, \ldots ,\mu_n)$$.

What should I do next? (Or is there another approach, not using the fact that $S, T$ are simultaneously diagonalizable?)


1 Answer 1


There is another approach.

It is a classical theorem such that an operator is diagonalizable if and only if its minimal polynomial has no multiple roots. It can be proved using Jordan form, Primary Decomposition theorem, Rational decomposition theorem, whatever you like.

Now suppose $m(x)$ is the minimal polynomial for $T$. Then, $m(T|_{V_\lambda})=0$ so that the minimal polynomial $m_1(x)$ of $T|_{V_\lambda}$ divides $m(x)$. Since $m$ has no multiple roots, $m_1$ doesn't have multiple roots, too, so that $T|_{V_\lambda}$ is diagonalizable.


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