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**Anthonny** 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.

Tonio

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.

Thank you.