A function evaluated at a operator If dim$(V)=n$ and $\tau \in \mathcal {L}(V)$, prove that there is a nonzero polynomial $p(x)\in F[x]$ for which $p(\tau)=0$. I don't understand how come a function can be evaluated at a linear operator $\tau$.
 A: Hint: $\mathcal{L}(V)$ is also a finite dimensional vector space (over what?) Thus, you can find some integer $m$ (you can even explicitly find $m$ in terms of $n$) such that:
$$\{1,\tau, \tau^2,\dots \tau^m\}$$ are linearly dependent. I think you can continue from here.
A: It's not a function, it's a polynomial. For polynomials, it's easy to understand:
if the polynomial is $f(x) = a_0 + a_1 x + \ldots + a_n x^n$, and the operator is $\tau$,
then $f(\tau) = a_0 I + a_1 \tau + \ldots + a_n \tau^n$.
This works in any algebra over a field $F$, where the polynomial has coefficients in $F$.
A: First, how to evaluate a polynomial $p(x) \in F[x]$ at a linear operator $\tau \in \mathcal L(V)$?  The first thing to realize is that powers of $\tau$ make sense as elements of $\mathcal L(V)$; e.g., for
$\vec v \in V, \tag 1$ we have
$\tau \vec v \in V, \tag 2$
so it makes sense to take
$\tau^2 \vec v = \tau(\tau \vec v); \tag 3$
indeed, accepting (3), we may inductively define
$\tau^{k + 1} \vec v = \tau (\tau^k\vec v); \tag 4$
thus we define $\tau^m$ for any $m \in \Bbb N$; if $\tau$ is represented by a matrix
$[\tau_{ij}], \tag 5$
then it is easy to see that $\tau^m$ is represented by the matrix
$[(\tau^m)_{ij}] = [\tau_{ij}]^m. \tag 6$
In any event, given $\tau^m$, we see that
$(\alpha \tau^m)(\vec v) = \alpha (\tau^m \vec v); \tag 7$
and we may also construct and evaluate expressions such as
$(\alpha \tau^l + \beta \tau^m)\vec v =  \alpha \tau^l \vec v + \beta \tau^m \vec v \tag 8$
using the ordinary linearity properties of $\mathcal L(V)$; continuing in this vein, we can for
$p(x) \in F[x], \; p(x) = \displaystyle \sum_0^m p_i x^i, \tag 9$
meaningfully define
$p(\tau) = \displaystyle \sum_0^m p_i \tau^i \in \mathcal L(V), \tag{10}$
which is the general formula for $p(\tau)$.
Now we recall that
$\dim_F \mathcal L(V) = n^2; \tag{11}$
thus the set
$\{I, \tau, \tau^2, \ldots, \tau^{n - 1}, \tau^{n^2 - 1}, \tau^{n^2} \} \subset \mathcal L(V), \tag{12}$
having as it does $n^2 + 1$ elements, must be linearly dependent over $F$; hence we may find $n^2 +1$ members
$c_i \in F,  \tag{13}$
not all $0$, such that
$\displaystyle \sum_0^{n^2} c_i \tau_i = 0; \tag{14}$
$\tau$ then satisfies the polynomial
$c(x) = \displaystyle \sum_0^{n^2} c_i x^i \in F[x]. \tag{14}$
We have shown that every $\tau \in F[x]$ satisfies a polynomial of degree at most $n^2$, but this is far from optimal.   For example, the well-known Cayley-Hamilton theorem shows that $\tau$ obeys a degree-$n$ polynomial; but the proof of that result is somewhat more involved than what we have done here, and hence will be saved for yet another post.
A: Let $p$ be the characterstic Polynomial of $\tau$. Then there exist $a_0,a_1,\cdots a_{n-1}\in\mathbb{C}$ such that $\det(\tau-\lambda I)=p(t)=a_0+a_1 t+\cdots a_{n-1}t^{n-1}=0$. Plugging $\tau$ yeilds
$p(\tau)=0$
