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 $C[0,1]$ be the linear space of continuous functions on $[0,1]$; this linear space is a complete normed linear space (i.e., a Banach Space) when given the max norm $\|f\|=\max_{t \in [0,1]}|f(t)|$. The Volterra operator
$$
(Vf)(x) = \int_{0}^{x}f(t)dt
$$
maps continuous functions to continuous functions. $V$ has trivial null space because $(Vf)(x)=0$ for all $x$ implies $f(x)=0$ for all $x$ by the Fundamental Theorem of Calculus. So $V$ does not have $0$ as an eigenvalue. If $Vf=\lambda f$ for a non-zero $\lambda$, then $f=\frac{1}{\lambda}Vf$ is continuously differentiable with $f'=\frac{1}{\lambda} f$ and $f(0)=0$, which has only the trivial solution $f\equiv 0$.
Best Answer
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.