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

Question

Let $f:V \to V$ be a linear map such that $(f\circ f)(v) = -v$. Prove that if $V$ is a finite dimensional vector space over $\mathbb R$, $V$ is even dimensional.

From what I can figure out for myself, if $V$ is finite dimensional, then every basis of $V$ is finite, i.e. a linearly independently subset of $V$ has a finite number of vectors.

And I figure that if $V$ is even dimensional, then every basis of $V$ is even, i.e. a linearly independently subset of $V$ has an even number of vectors.

But I'm not sure how to connect these two points.

Answer 1

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.]

Answer 2

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 $\mathbf R$.

Answer 3

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.

Answer 4

Just for fun...

As in Callus's answer, observe that if $f(f(x))=-x$ for all $x\in\mathbb R^n$ then there is no $x\ne 0$ such that $f(x)=\lambda x$ for some $\lambda\in\mathbb R$ (since then we would have $\lambda^2x=-x$, which is impossible). We may define a map $F\colon\mathbb R^n\to\mathbb R^n$ by $$ F(x) = \left<x,x\right>f(x) - \left<x,f(x)\right>x $$ which is a scalar multiple of the usual projection map of $f(x)$ on to the plane tangent to $x$. This map has no zeroes other than $x=0$; indeed, if $x\ne 0$ and $F(x)=0$, it means that $$ f(x) = \frac{\left<x,f(x)\right>}{\left<x,x\right>}x $$ which we already know to be impossible unless $x=0$. Observing that $\left<x,F(x)\right>=0$ for all $x\in\mathbb R^n$, we see that $F$ restricts to a smooth tangent field on $S^{n-1}$ with no zeroes; by the hairy ball theorem, it follows that $n-1$ is odd, and so $n$ is even.

Answer 5

Since $f^2 = -I$, we see that the minimal polynomial of $f$ over $\mathbb{R}$ is $x^2 + 1 \in \mathbb{R}[x]$. The minimal polynomial and characteristic polynomial share their root sets, so the characteristic polynomial of $f$ must be $(x^2 + 1)^k$ for some $k\ge 1$. But the degree $2k$ of the characteristic polynomial is the dimension of $V$, hence $V$ is even-dimensional.

Answer 6

Use induction on the dimension. For non zero $v$, $\{v, f(v)\}$ is linear independent and spans a 2d subspace closed under f. Then f is well defined on the quotient $V/\{v,f(v)\}$, which has dimension 2 less. (tyro1's answer is I think along these lines but a bit garbled).