Return to Answer

Essentially the same as Marcos Cossarini's proof, but without Cauchy-Schwarz.

Let $A\in R^{n\times n}$ symmetric of rank $\ge 1$. By compactness let $x\in R^n$ with $\|x\|=1$ be such that $\|Ax \| = \max_{u\in R^n: \|u\|=1} \|Au\|$ and set $y = Ax/\lambda$ for $\lambda=\|Ax\|>0$. Then \begin{align} \|A(x+ y) - \lambda(x+y)\|^2 &=\| Ay - \lambda x\|^2 \\&= \|A y\|^2 + \lambda^2 - 2 \lambda y^TA x \\&= \|A y\|^2 + \lambda^2 - 2\lambda^2 \\&\le 0. \end{align} Either $x+y\ne 0$ is eigenvector for $\lambda$ or $x+Ax/\lambda=$ and $x$ is is eigenvector for $-\lambda$.

Essentially the same as Marcos Cossarini's proof, but without Cauchy-Schwarz.

Let $A\in R^{n\times n}$ symmetric of rank $\ge 1$. By compactness let $x\in R^n$ with $\|x\|=1$ be such that $\|Ax \| = \max_{u\in R^n: \|u\|=1} \|Au\|$ and set $y = Ax/\lambda$ for $\lambda=\|Ax\|>0$. Then \begin{align} \|A(x+ y) - \lambda(x+y)\|^2 &=\| Ay - \lambda x\|^2 \\&= \|A y\|^2 + \lambda^2 - 2 \lambda y^TA x \\&= \|A y\|^2 + \lambda^2 - 2\lambda^2 \\&\le 0. \end{align} Either $x+y\ne 0$ is eigenvector for $\lambda$ or $x+Ax/\lambda=0$ and $x$ is is eigenvector for $-\lambda$.

Essentially the same as Marcos Cossarini's proof, but without Cauchy-Schwarz.

Let $A\in R^{n\times n}$ symmetric of rank $\ge 1$. By compactness let $x\in R^n$ with $\|x\|=1$ be such that $\|Ax \| = \max_{u\in R^n: \|u\|=1} \|Au\|$ and set $y = Ax/\lambda$ for $\lambda=\|Ax\|>0$. Then \begin{align} \|A(x+ y) - \lambda(x+y)\|^2 &=\| Ay - \lambda x\|^2 \\&= \|A y\|^2 + \lambda^2 - 2 \lambda y^TA x \\&= \|A y\|^2 + \lambda^2 - 2\lambda^2 \\&\le 0. \end{align} Either $x+y\ne 0$ is eigenvector for $\lambda$ or $x=-y=-Ax/\lambda$ is eigenvector for $-\lambda$.

Essentially the same as Marcos Cossarini's proof, but without Cauchy-Schwarz.

Let $A\in R^{n\times n}$ symmetric of rank $\ge 1$. By compactness let $x\in R^n$ with $\|x\|=1$ be such that $\|Ax \| = \max_{u\in R^n: \|u\|=1} \|Au\|$ and set $y = Ax/\lambda$ for $\lambda=\|Ax\|>0$. Then \begin{align} \|A(x+ y) - \lambda(x+y)\|^2 &=\| Ay - \lambda x\|^2 \\&= \|A y\|^2 + \lambda^2 - 2 \lambda y^TA x \\&= \|A y\|^2 + \lambda^2 - 2\lambda^2 \\&\le 0. \end{align} Either $x+y\ne 0$ is eigenvector for $\lambda$ or $x+Ax/\lambda=$ and $x$ is is eigenvector for $-\lambda$.

Essentially the same as Marcos Cossarini's proof, but without Cauchy-Schwarz.

Let $A\in R^{n\times n}$ symmetric of rank $\ge 1$. By compactness let $x\in R^n$ with $\|x\|=1$ be such that $\|Ax \| = \max_{u\in R^n: \|u\|=1} \|Au\|$ and set $y = Ax/\|Ax\|$. Set $\epsilon=+1$ if $y^Tx \ge 0$ and $\epsilon=-1$ if $y^Tx<0$. Then $\|x+\epsilon y\|^2 \ge 2$ so that $x+\epsilon y \ne 0$. With $\lambda=\epsilon \|Ax\|$ we have $Ax = \epsilon \lambda y$ and \begin{align} \|A(x+\epsilon y) - \lambda(x+\epsilon y)\|^2 &=\|\epsilon Ay - \lambda x\|^2 \\&= \|A y\|^2 + \lambda^2 - 2 \epsilon \lambda y^TA x \\&= \|A y\|^2 + \lambda^2 - 2\lambda^2 \\&\le 0 \end{align} since $ \|Ay\|\le \|Ax\|=|\lambda|$ by definition of $x$.

Essentially the same as Marcos Cossarini's proof, but without Cauchy-Schwarz.

Let $A\in R^{n\times n}$ symmetric of rank $\ge 1$. By compactness let $x\in R^n$ with $\|x\|=1$ be such that $\|Ax \| = \max_{u\in R^n: \|u\|=1} \|Au\|$ and set $y = Ax/\lambda$ for $\lambda=\|Ax\|>0$. Then \begin{align} \|A(x+ y) - \lambda(x+y)\|^2 &=\| Ay - \lambda x\|^2 \\&= \|A y\|^2 + \lambda^2 - 2 \lambda y^TA x \\&= \|A y\|^2 + \lambda^2 - 2\lambda^2 \\&\le 0. \end{align} Either $x+y\ne 0$ is eigenvector for $\lambda$ or $x=-y=-Ax/\lambda$ is eigenvector for $-\lambda$.