.I'm having trouble seeing how induction is used in this proof for this theorem which states that every operator on an odd-dimensional real vector space has an eigenvalue.
Hint (the size of an elephant): every real polynomial of odd degree has a real root.
Oh, and no induction at all needed.
The proof starts by supposing V is a real vector space with odd dimension. Then the proof states that it will be using induction in step sizes of 2 on the dimension of V (which I understand since we're proving this for odd dimensions).
The induction starts with when dimV=1 then the operator on V has an eigenvalue which I understand.
Then the proof supposes that dimV is an odd number greater than 1. Using induction we can assume that the desired result holds for dimensions 2 less than dimV.
I can't clearly see how this is true i.e. how, say, if dimV=3 then the operator on V has an eigenvalue.