Lemma: let
$$V_{\lambda}^{m}(A) = \{v\in V|(A - \lambda E)^{m}v = 0\}$$
For any $A\in\{A_1, A_2, \ldots\}$. Then for any $B\in\{A_1, A_2, \ldots\}$
$$B(V_{\lambda}^{m}(A))\subset V_{\lambda}^{m}(A).$$
Proof: if $v\in V_{\lambda}^{m}(A)$ then $(A - \lambda E)^{m}Bv = B(A - \lambda E)^{m}v = 0.$ QED
Induction on the dimension:
Consider two special cases:
First: there is operator $A\in\{A_1, A_2, \ldots\}$ with tho different eigenvalues $\lambda_1$ and $\lambda_2$. Then subspace
$$W = V_{\lambda_1}^{n}(A) \subsetneq V$$
is invariant under the action of any operator $A_i$ (by lemma) and
$$\dim W < \dim V.$$
So by induction $A_1, A_2, \ldots$ have common eigenvector in $W$.
Second: all eigenvalues of any operator $A\in\{A_1, A_2, \ldots\}$ are the same. This case splits into two:
1. There is operator $A\in\{A_1, A_2, \ldots\}$ such that
$$V_{\lambda}^{1}(A) \neq V.$$
Then $V_{\lambda}^{1}(A)$ is invariant under the action of any operator $A_i$ and
$$\dim V_{\lambda}^{1}(A) < \dim V.$$
So by induction $A_1, A_2, \ldots$ have common eigenvector in $V_{\lambda}^{1}(A)$.
2. For any $A\in\{A_1, A_2, \ldots\}$
$$V_{\lambda}^{1}(A) = V.$$
Then any vector of $V$ is common eigenvector of operators $A_1, A_2, \ldots$. QED
Can you check my solution?
Thanks a lot!
Let $n = \dim V$.
Choose $v \in V$ with $v \ne 0$. Then
$$
v, Tv,T^2 v, \dots , T^n v
$$
is not linearly independent, because $V$ has dimension $n$ and we have
$n+1$ vectors. Thus there exist complex numbers $a_0, \dots , a_n$, not all
$0$, such that
$$
0 = a_0 v + a_1 Tv + \dots + a_n T^n v.
$$
Note that $a_1, \dots, a_n$ cannot all be $0$, because otherwise the equation above would become $0 = a_0 v$, which would force $a_0$ also to be $0$.
Make the $a$'s the coefficients of a polynomial, which by the Fundamental Theorem of Algebra has a factorization
$$
a_0 + a_1 z + \dots + a_n z^n = c (z - \lambda_1) \dotsm (z - \lambda_m),
$$
where $c$ is a nonzero complex number, each $\lambda_j \in \mathbf{C}$, and
the equation holds for all $z \in \mathbf{C}$ (here $m$ is not necessarily equal to $n$, because $a_n$ may equal $0$). We then have
\begin{align*}
0 &= a_0 v + a_1 Tv + \dots + a_n T^n v \\
&=(a_0 I + a_1 T + \dots + a_n T^n) v\\
&= c(T - \lambda_1 I) \dotsm (T - \lambda_m I)v.
\end{align*}
Thus $T - \lambda_j I$ is not injective for at least one $j$. In
other words, $T$ has an eigenvalue.
Best Answer
Hint: Consider $$A=\begin{pmatrix}1&0\\0&-1\end{pmatrix}. $$ Or $$A=\begin{pmatrix}0&1\\0&0\end{pmatrix}. $$