Let $V$ be a real vector space of dimension $n$ and let $\langle \, \cdot\, , \,\cdot\, \rangle$ be an inner product on $V$. We can define a linear functional on the space of endomorphisms of $V$ by $$ \alpha(A) := \int_{S^{n-1}} \langle Av, v \rangle d\mu, $$ where $S^{n-1}$ is the unit sphere defined by the inner product and $d\mu$ is the Lebesgue measure on $S^{n-1}$. This functional is actually a multiple of the trace. Here's a nasty proof of this:
Let $(e_1,\ldots,e_n)$ be an orthonormal basis of $V$. Given $v \in V$ we write $v = v_1 e_1 + \cdots + v_n e_n$, so $v \in S^{n-1}$ if and only if $\sum |v_i|^2 = 1$. We now write $$ Av = \sum_{j=1}^n v_j \, A e_j = \sum_{j,k=1}^n v_j \, a_{jk} e_k $$ so that $$ \langle Av ,v \rangle = \sum_{j,k} a_{jk} \, v_jv_k. $$ It's now classically known that $\int_{S^{n-1}} v_j v_k d\mu = c \, \delta_{jk}$ for some constant $c$ that's not terribly important here (and depends on the normalization of $\mu$ anyway), so we get $$ \alpha(A) = \int_{S^{n-1}} \langle Av, v \rangle d\mu = c \operatorname{tr}(A). $$ Question Is there not a better way of doing this? That is, is there not some way of seeing that $\alpha = c \operatorname{tr}$ for some nonzero $c$ without breaking out the local coordinates? This basically comes down to showing that $\alpha(AB) = \alpha(BA)$ for any $A,B \in \operatorname{End} V$, but I can't see how to show that (or equivalently that $\alpha$ vanishes on the commutator) without going through the same calculations as above.
