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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 $tau$.

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

$tau^{k + 1} vec v = tau (tau^kvec v); tag 4$

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.

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.

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 $F$.

Let $p$ be the characterstic Polynomial of $tau$. Then there exist $a_0,a_1,cdots a_{n-1}inmathbb{C}$ such that $det(tau-lambda I)=p(t)=a_0+a_1 t+cdots a_{n-1}t^{n-1}=0$. Plugging $tau$ yeilds
$p(tau)=0$