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Theorem Every operator on an odd-dimensional real vector space has an eigenvalue.

Incomplete proof Suppose that V is a real vector space with odd dimension. The result is obviously true if dim V=1. So now we assume dimension of V is an odd number greater than 1. Assume that the result is true for all real vector spaces with dimension 2 less than dim V. Now we need to prove that T has an eigenvalue. If it does , our work is done. If not, there is a two-dimensional subspace U of V that is invariant under T.

(Doubt) I am aware that every operator on a finite dimensional vector space has an invariant subspace of dimension 1or2. So why not consider two subspaces U and W of dimension 1 and 1 respectively. How we are sure that there will exist a 2 dimensional invariant subspace?

After that proof is very much clear to me.

NB This proof is from "Linear algebra done right'' by Sheldon Axler ,2nd edition chapter 5, last section.

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Because if there's an invariant subspace of dimension $1$, any nonzero element of that subspace is an eigenvector and we've assumed $T$ doesn't have any eigenvectors.

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  • $\begingroup$ Why can't same thing happen in two dimensional space? $\endgroup$ – blue boy Mar 12 at 6:16
  • $\begingroup$ Oh..i think i get it. If T has an eigenvalue (say è) then T-èI is not injective . And hence one dimensional invariant subspace . $\endgroup$ – blue boy Mar 12 at 6:22

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