Let $L\in End(V)$ with $L(V)=W$. Then $Tr(L)=Tr(L|_W)$ Let $V$ be a finite dimensional vector space and $W$ a subspace. Let $L$ be an endomorphism with image is $W$. Then $Tr(L)=Tr(L|_W)$ where $L|_W$ denotes the restriction of $L$ to $W$.
I am unsure what definition of trace I should use to prove this, it seems  "obvious" if the trace is simply the sum of the diagonal elements of a matrix. But I am not sure if we are allowed to restrict linear maps when using such a definition.
One idea is: $L|_W$ is an isomorphism of $W$, this gives is a nice basis of $W$ which we can extend to a basis of $V$ such that the matrix of $L$ is 0 in the last $n-m$ collums and scalars along the first m diagonals. Then maybe $L|_W$ is given as matrix just by getting rid of the last $n-m$ columns and then we have the result.
Edit:
Specifically I am confused with how to calculate the trace of the restriction of a map. I know that the trace is the sum of the diagonals of the matrix that represents the map, and the this does not depend on the basis that we choose. But how do we do this when we restrict our map to a subspace? It cannot be that we still sum over the diagonals. Suppose that was the case then we have $0=Tr(L|_{Ker(L)}=Tr(L)\neq 0$.
 A: This is a special case of the general result that if $W$ is an invariant subspace for $L \in \operatorname{End}(V)$ (i.e. $L(W) \subseteq W$), then we can form a restriction map $L {\mid}_W \in \operatorname{End}(W)$ and an induced map $\bar L \in \operatorname{End}(V / W)$; and:
$$\operatorname{Tr}(L) = \operatorname{Tr}(L {\mid}_W) + \operatorname{Tr}(\bar L).$$
In the particular case given here, we certainly have that $L(W) \subseteq L(V) \subseteq W$, so $W$ is an invariant subspace for $L$.  Also, the induced map $\bar L$ is equal to 0 in this case since $\bar L(x + W) = (Lx) + W = 0 + W$ for every $x \in V$.  Therefore, $\operatorname{Tr}(\bar L) = 0$, and the desired result follows.

The idea for the proof of the general result is: start with a basis $\beta_W = \{ x_1, \ldots, x_m \}$ of $W$, and extend to a basis $\beta_V = \{ x_1, \ldots, x_m, x_{m+1}, \ldots, x_n \}$ of $V$.  Then in $[L]_{\beta_V}$, the upper left $m \times m$ submatrix is equal to $[L {\mid}_W]_{\beta_W}$, and the lower right $(n-m) \times (n-m)$ submatrix is equal to $[\bar L]_{\beta_{V / W}}$ where $\beta_{V / W} := \{ x_{m+1} + W, \ldots, x_n + W \}$ is a basis for $V / W$.
