Timeline for real symmetric matrix has real eigenvalues - elementary proof

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Nov 17, 2015 at 8:53	comment	added	Pietro Majer		I usually start with the remark: the Rayleigh ratio naturally comes in the eigenvalue problem of a symmetric matrix, because if $Au=\lambda u$, then of course $\lambda =R(u)$. Also, differentiating, $$\nabla R(x)= 2\|x\|^{-2}\big( Ax - R(x)x \big).$$ So critical points of $R$ are exactly eigenvectors of $A$, and the corresponding critical levels are exactly the associated eigenvalues.
Jan 17, 2013 at 21:20	vote	accept	marjeta
Jan 14, 2013 at 21:31	comment	added	Alexandre Eremenko		Marcos: yes. ACL's explanation is one way to do it.
Jan 14, 2013 at 0:15	comment	added	ACL		Marcos. In that case, you can prove the Lagrange multiplier relation hand: if $\lambda$ is the minimum, then for every $y$, $(x+y)^TA(x+y)\geq \lambda (x+y)^T(x+y)$, that is, $2((y^T (Ax-\lambda x)\geq \lambda y^Ty - y^TAy$. The LHS is homoegeneous of degree $1$, the RHS of degree $2$. So the LHS has to be zero for every $y$. This implies $Ax=\lambda x$.
Jan 13, 2013 at 23:56	comment	added	Marcos Cossarini		Alexander, when you said that the minimum is an eigenvalue, did you mean to prove it by applying the Lagrange multiplier equation to the function $f(x)=x^tAx$ restricted to a level set of $g(x)=x^tx$, or did you have a different idea in mind?
Jan 11, 2013 at 23:32	vote	accept	marjeta
Jan 11, 2013 at 23:33
Jan 11, 2013 at 14:06	history	answered	Alexandre Eremenko	CC BY-SA 3.0