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Is there any proof of the fact that the eigenvalues of symmetric matrices (i.e. $A\in\mathbb{R}^{n\times n}$ with $A^t=A$) are real without the use of the concept of complex numbers?

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  • $\begingroup$ Cauchy Interlace Theorem (en.wikipedia.org/wiki/Min-max_theorem) is a chance, but it deeply relies on the existence of an orthonormal base for the dot product given by a symmetric matrix. $\endgroup$ – Jack D'Aurizio Jan 29 '15 at 12:21
  • $\begingroup$ What is your definition of eigenvalue? It either uses complex numbers, or it implies that every eigenvalue is real. $\endgroup$ – daw Jan 29 '15 at 12:52
  • $\begingroup$ Not sure if it helps, but (assuming $Au = \lambda u$) by expanding $\lambda ^2 \| u \|^2$ it is not too difficult to see that $\lambda ^2 \ge 0$. At least there is no need to use conjugates if that's what you wanted to avoid. $\endgroup$ – flavio Jan 29 '15 at 13:00
  • $\begingroup$ What's wrong in using complex numbers? $\endgroup$ – egreg Jan 29 '15 at 14:09
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Let $\rho=\max_{\|x\|=1}\|Ax\|$ and $u=\arg\max_{\|x\|=1}\|Ax\|$. Then $Au=\rho v$ for some unit vector $v$ and $\|Av\|\le\rho$. But we also have $\langle Av,u\rangle=\langle v,A^Tu\rangle=\langle v,Au\rangle=\rho$. Hence $Av=\rho u$. Therefore, either $(\rho,u)$ (when $u=v$) or $(-\rho,u-v)$ (when $u\ne v$) is an eigenpair of $A$. Since the orthogonal complement of this eigenspace is invariant under $A$ (because $A$ is symmetric), we have reduced the dimension of the problem by $1$. Proceed recursively, $A$ has $n$ real eigenvalues and an orthogonal eigenbasis.

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