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5

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.

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Just a note: you say "then the basis", but this doesn't make sense. There isn't a unique basis. – Zachary Selk 5 hours ago
    
Oh, right - thanks for pointing that out :) Would it be correct if I just changed it to "a basis" rather than "the basis"? – maths123 5 hours ago
    
I would say "then every basis is finite" (not finite dimensional, a basis doesn't have a dimension, a vector space does). – Zachary Selk 5 hours ago
3  
(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 5 hours ago
1  
@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. – amcerbu 4 hours ago
up vote 9 down vote accepted

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. 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.]

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Great, that makes sense! Thank you :) – maths123 4 hours ago
    
@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. – Mathemagician1234 4 hours ago

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