Prove that if V is finite dimensional then V is even dimensional?

I heard this story from David Lieberman:

Once this question was included in the Qual (qualifying exam) for Harvard graduate students. As it turned out this one question perfectly predicted all students' performance, so the exam's other 17 questions were not necessary! Indeed:

• Every student who did not solve this question flunked their Qual.

• Every student who solved this problem by fiddling with Jordan canonical forms and the like got a "conditional pass".

• Every student who solved this problem using the determinant passed.

$(\det f)^2 = \det (f \circ f) = \det -I = (-1)^n$ where $n = \dim V$, so $(-1)^n \geq 0$ and $n$ is even.

[Note: This assumes that the matrix has real coefficients. As noted in the comments, the result would be wrong over the complex numbers (e.g., let $n=1$ and $f=i$) and some other fields.]

Here is another solution. We can put the structure of a complex vector space on $V$ by defining $$ (a+bi)\cdot v=av+bf(v) $$ for all $v\in V$ and for all $a,b\in\mathbf R$. This works exactly since $f^2=-1$. Then $V$ is of course a finite dimensional vector space over $\mathbf C$, and therefore of even dimension over $\mathbf R$.

This is kind of similar to Noam's answer, but...

You could also argue that there are no eigenvectors with real eigenvalues, since if $f(v)=\lambda v$, then $-v=f(f(v))=\lambda^2 v$, so $\lambda^2 = -1$. However, if $V$ is odd dimensional, then the characteristic polynomial is odd degree, and therefore has a real root, and therefore $f$ has a real eigenvalue.