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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 an odd dimension.

The result is obviously true if $\dim V=1$. So now we assume the dimension of $V$ is an odd number greater than $1$. Assume that the result is true for all real vector spaces with dimension equaling $\dim V-2$.

Now we need to prove that $T$ has an eigenvalue. If it does, our work is done. If not, there is a $"-D$ subspace $U\leqslant V$ that is invariant under $T$.

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

After that proof is very much clear to me.

NB

Source: Alexander, Sheldon, 'Linear algebra done right', $2^{\text{nd}}$ 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, 2019 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, 2019 at 6:22

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