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

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.

• Just a note: you say "then the basis", but this doesn't make sense. There isn't a unique basis.
– user223391
Commented Jun 17, 2016 at 3:59
• Oh, right - thanks for pointing that out :) Would it be correct if I just changed it to "a basis" rather than "the basis"? Commented Jun 17, 2016 at 4:00
• I would say "then every basis is finite" (not finite dimensional, a basis doesn't have a dimension, a vector space does).
– user223391
Commented Jun 17, 2016 at 4:04
• (1) I'm not sure there's much to connect just yet. So far we've ignored $f$. (2) What field are you working over? I think that matters here.
– Hoot
Commented Jun 17, 2016 at 4:15
• @Hoot is right, the field is important. If we're working over $\mathbb F_5$ as a 1-dimensional vector space, $v \mapsto 2v$ has the desired properties but our space has odd dimension. Commented Jun 17, 2016 at 4:35

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.]

• Great, that makes sense! Thank you :) Commented Jun 17, 2016 at 5:05
• @NoamD.Elkies I just got it and now I'm kicking myself.I'm surprised they didn't have to put those graduate students on suicide watch after that. Commented Jun 17, 2016 at 5:10
• That must have been in the quaint old days, if out of 17 graduate students none observed that f(v) is the data of a complex structure on V.
– zyx
Commented Jun 18, 2016 at 3:30
• This story goes back 30+ years. Indeed nowadays graduate students and even the top undergraduates come with much stronger backgrounds than we could expect a generation ago. The Internet probably deserves a good part of the credit for this... Commented Jun 18, 2016 at 18:48
• @NoamD.Elkies Can confirm, just graduated and view the internet (and this website!) as being highly responsible for my development of mathematics both as a skill and as an interest. Commented Apr 4, 2017 at 17:31

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 $\mathbf R$.

• Sheldon "use $\det$ only as a last resort" Axler would be proud. +1 Commented Jun 17, 2016 at 18:25

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.

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.

• I too solved it the same way! Commented May 17, 2020 at 7:28

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.

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).