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I got stuck trying to show that if $A,B$ are two linear operators on a finite dimensional vector space $V$ over the field $\mathbb{F}$, then $A$ and $B$ share an eigenvector. Quick Googling revealed that main idea in the proof is that each eigenspace of $A$ (resp. of $B$) is invariant under $B$ (resp. $A$) and therefore the restriction of $B$ to the union of the eigenspaces of $A$ has an eigenvector. What I don't know and don't know how to look up more about it is the connection between an invariant subspace $U$ under a linear mapping $B$ of the ambient space $V$ and $B$'s eigenvectors. So what, if any, connection is there?

Cartesian Bear
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  • @user977780 It does not answer my question. Answers to that post either discuss why the asked question is not true or they themselves use it as a fact that a linear operator has an eigenvalue in an invariant space. – Cartesian Bear May 11 '23 at 11:55
  • The result is not true for a general field. I have edited my answer with a counter-example. – geetha290krm May 11 '23 at 12:13

2 Answers2

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If $\mathbb F=\mathbb R$, $A=I$ and $B$ is rotation by $90^{0}$ then $A$ and $B$ commute but $B$ has no eigen vectors, so the result is false for a general $\mathbb F$.

If $\mathbb F=\mathbb C$ we can argue as follows: Let $M=\{x: Ax=\lambda x \,\, \text {for some} \,\,\lambda\}$. If $x \in M$ then $Ax=\lambda x$ for some $\lambda$. Now consider $Bx$. Let us show that $Bx \in M$. We have $A(Bx)=B(Ax)=B(\lambda x)=\lambda Bx$ which proves that $Bx \in M$. So we can consider $B$ as an operator from $M$ into $M$ (by restriction). This new operator has an eigen vector. This finishes the proof.

geetha290krm
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  • I still don't understand how $B$'s restriction has an eigenvector. You showed that for any $x\in M:(AB)(x)=\lambda B(x)$ and $B(M)\subset M$. Why is this sufficient to conclude that $B\mid_M$ has an eigenvector? – Cartesian Bear May 11 '23 at 11:53
  • @CartesianBear I have edited my answer. – geetha290krm May 11 '23 at 12:13
  • Thanks for the updated version. But still, and this could be just me forgetting some basic result from linear algebra, but why does $B\mid_M$ have an eigenvector? $M\subset V$ is a subspace of our ambient space and $B\mid_M$ is a linear operator on $M$. Why does $B\mid_M$ have an eigenvector? Existence of the eigenvalues of $B$ on $V$ is given by the characteristic polynomial. Is it the same thing in this subspace case? – Cartesian Bear May 11 '23 at 12:20
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    That is just basic linear algebra. If $T$ is any LT on a finite dimesional vector space $W$ over $\mathbb C$ then the polynomial equation $det(T-\lambda I)=0$ has at least one root and for a root $\lambda$ of this equation there exists $x \neq 0$ such that $Tx=\lambda x$. Take $W=M$ and $T$ to be the restriction of $B$ to $M$. @CartesianBear – geetha290krm May 11 '23 at 12:27
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Let $Ax = \lambda x$, then $ABx = BAx = \lambda Bx$ so $Bx$ is an eigenvector of $A$. Now suppose that $\{u_1,\ldots,u_k\}$ forms a basis for the eigenspace of $A$ corresponding to $\lambda$. Then $Bu_i = \sum_{j=1}^k p_{ji}u_j$ for some $p_{ji}\in \mathbb{F}, i,j=1,\ldots,k$.

Finally, form matrix the $P = (p_{ji})_{j,i=1}^k$. If $(\mu,\xi)$ is an eigenpair of $P$, then set $q = \sum_{i=1}^k \xi_i u_i$ and note that $ Aq = \lambda q $ and $$ Bq= B\sum_{i=1}^k \xi_i u_i= \sum_{i=1}^k \xi_i \sum_{j=1}^k p_{ji} u_j = \sum_{j=1}^ku_j \sum_{i=1}^k p_{ji}\xi_i = \sum_{j=1}^k u_j (P\xi)_{j}=\mu \sum_{j=1}^k u_j \xi_j = \mu q $$ so $q$ is an eigenvector of $A$ and $B$.

V.S.e.H.
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