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Is there a determinant-free proof that $\textbf{A} \in \mathbb{R}^{n \times n}$ must have at least one real eigenvalue when $n$ is odd?

I have seen a few other definitions of "the set of eigenvalues" which do not invoke the determinant, specifically the complement of the resolvent set, or the set of points for which $\lambda\textbf{I}_{n \times n} - \textbf{A}$ is singular.

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  • $\begingroup$ What do you mean by "for which $\lambda I-A$ is defined"? $\endgroup$
    – Vim
    Sep 18, 2017 at 1:18
  • $\begingroup$ I meant singular, I have fixed it. $\endgroup$ Sep 18, 2017 at 1:19
  • $\begingroup$ The characteristic polynomial is of odd degree and therefore has at least one real root - perhaps? $\endgroup$
    – NickD
    Sep 18, 2017 at 1:20
  • $\begingroup$ The characteristic polynomial as I know it comes from a determinant. Though, through some searching around I found this paper which is a little bit interesting: jstor.org/stable/24338320?seq=3#page_scan_tab_contents $\endgroup$ Sep 18, 2017 at 1:20
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    $\begingroup$ Axler's Down with Determinants has a proof (the proof of existence of a complex eigenvalue is easy; proving there is a real one requires more knowledge of the structure of the eigendecomposition, and occurs quite late in the paper: Thm 8.2). $\endgroup$
    – Chappers
    Sep 18, 2017 at 2:07

2 Answers 2

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This is a consequence of the hairy ball theorem.

Given a real $n \times n$ matrix $A$, define a function on the unit sphere in $\mathbb R^n$ by mapping a unit vector $\vec x$ to the component of $A\vec x$ perpendicular to $\vec x$: that is, $$\vec x \mapsto A \vec x - \langle A\vec x, \vec x\rangle \vec x.$$ Being perpendicular to $\vec x$, the result is always tangent to the sphere.

When $n$ is odd, the unit sphere in $\mathbb R^n$ is even-dimensional, and so (by the hairy ball theorem) there is no nonvanishing continuous function with the above property. This means there must be some unit vector $\vec x \in \mathbb R^n$ such that $A \vec x - \langle A\vec x, \vec x\rangle \vec x = \vec 0$: that is, $\vec x$ is an eigenvector of $A$ with eigenvalue $\langle A\vec x, \vec x\rangle$.

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  • $\begingroup$ Axler's Down with Determinants proved te existence of eigenvalues without using determinants. $\endgroup$
    – Vim
    Sep 18, 2017 at 2:13
  • $\begingroup$ @Vim Yes, but how is your comment related to the answer here? $\endgroup$
    – user1551
    Sep 18, 2017 at 10:34
  • $\begingroup$ @user1551 I was thinking if eigenvalues are proved to exist then we could just use the other answer's proof which is simpler. $\endgroup$
    – Vim
    Sep 18, 2017 at 14:40
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If $A$ is a matrix and $$ Av=\lambda v $$ then we also have $$ A\overline{v}=\overline{A}\overline{v}=\overline{Av}=\overline{\lambda v}=\overline{\lambda}\overline{v} $$ where we use the equality $A=\overline{A}$ because $A$ has real entries. Thus the eigenvalues of $A$ come in complex conjugate pairs, so in particular if $A$ has $2k+1$ eigenvalues, at least $1$ must be real.

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  • $\begingroup$ For this to be a complete proof, we need to know that $A$ has any eigenvalues (real or complex) to begin with. This has a determinant-free proof as well (see this paper for example) but that is not obvious. $\endgroup$ Sep 18, 2017 at 1:30
  • $\begingroup$ Thanks for the addition @MishaLavrov . It is something that I have taken for granted for so long that I didn't even think about it, but you are right, that is important $\endgroup$
    – TomGrubb
    Sep 18, 2017 at 1:33

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