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Let $U$ be a finite dimensional vector space over $\mathbb{C}$ and let $T_1 : U → U,\dots , T_n : U → U$ be pairwise commuting linear maps.

I was to show that there exists a $v ∈ V$ that is an eigenvector of all $T_i$ The hint given to me was to first prove that the eigenspaces of $T_i$ are preserved by $T_j$, but I don't quite get it

Could anyone help please?

egreg
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JD19
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2 Answers2

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Here's a proof using induction on $\dim(V)$, based on this answer and noting Ted's comment.

The case where $\dim(V)=1$ is easy (verify). For the main induction step we use this lemma:

lemma: Let $T_1$ and $T_2$ be commuting linear maps. Any eigenspace $E_\lambda$ of $T_1$ is preserved under $T_2$, that is $T_2(E_\lambda)\subseteq E_\lambda$.

proof: Let $\lambda$ be an eigenvector of $T_1$ and $E_\lambda$ the corresponding eigenspace. For any $u\in E_\lambda$ we have: $$T_1T_2u=T_2T_1u=T_2\lambda u=\lambda T_2u$$ This implies $(\forall u\in E_\lambda)(T_2u\in E_\lambda)$ or $T_2(E_\lambda)\subseteq E_\lambda$. $\blacksquare$

We now prove the induction step; let $\dim(V)>1$. Eigenvalues of $T_i$ are roots of the function $p_i(\lambda)=\det(T_i-\lambda I)$. By Libneiz's rule we can see (verify) that $p_i(\lambda)$ is a polynomial of degree exactly $\dim(V)$ in $\lambda$ (called the characteristic polynomial of $T_i$). By the fundamental theorem of algebra, $p_i$ has $\dim(V)>1$ roots (counted with multiplicity) in $\mathbb{C}$.

If each $T_i$ has only a unique eigenvalue (in which case all $T_i$ are scalar multiples of identity) the proof is easy (verify). Suppose there is some $T_k$ that has more than one distinct eigenvalues. Pick any of its eigenvalues, say $\lambda_k$. We must have (verify) $\dim(E_{\lambda_k})>0$ and $\dim(E_{\lambda_k})<\dim(V)$.

We know by the lemma above that $E_{\lambda_k}$ is conserved by all $T_i$ for $i\neq k$ and therefore the restriction of $T_i$'s on $E_{\lambda_k}$ are in fact lower dimensional endomorphisms (from $E_{\lambda_k}$ to itself). By the induction assumption, since $\dim(E_{\lambda_k})<\dim(V)$, there exists some nonzero $v\in W$ such that $v$ is an eigenvector of all $T_i$ for $i\neq k$. But $v$ is also, by definition of $E_{\lambda_k}$, an eigenvector of $T_k$ which completes the proof.

Community
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amirkdv
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Take two commuting linear operators on $V$, $S$ and $T$, and let $\lambda$ be an eigenvalue of $T$, with $v \in V$ in $E_\lambda$. \begin{equation} T(S(v)) = S(T(v)) = S(\lambda v) = \lambda S(v) \end{equation} This implies $S(v) \in E_\lambda$. Applying $S$ to the whole eigenspace, we get that $S(E_\lambda) \subset E_\lambda$. Thus, commuting linear transformations preserve each others eigenspaces.

Now, we will consider a decomposition of $\mathbb{C}^n$ into an direct sum of the eigenspaces of $T_1$ (the choice of $1$ is arbitrary). $$\mathbb{C}^n = E_{\lambda_1} \oplus \cdots \oplus E_{\lambda_m}$$

As we have shown that the eigenspaces of $T_i$ are preserved by $T_j$, we can decompose each of the eigenspaces in the above decomposition by another one of the linear maps. We repeat this until we have exhausted the set of linear maps $\{T_1, \cdots, T_k\}$.

At last, we assemble the recursive decompositions of eigenspaces to form a flat decomposition of $\mathbb{C}^n$. Each resulting eigenspace is an eigenspace of each of the linear maps $T_1, \cdots, T_k$.

Jacob Errington
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