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Let be $ A\in \mathbb{R}^{2,2} $ a symmetrical and positive definite matrix with different eigenvalues $ \lambda_1>\lambda_2>0 $. Further we have the rayleigh quotient $$ \max_{x\neq 0}\frac{\overline{x}^T\cdot A\cdot x}{\overline{x}^T\cdot x}=\max_{\|x\|_2=1}\overline{x}^T\cdot A\cdot x $$. For how many points $ x\in \mathbb{R}^2 $ the maximum $$ \max_{\|x\|_2=1}\overline{x}^T\cdot A\cdot x $$ is reached?

My idea: Let $$ A:=\begin{pmatrix}a&b\\b&c\end{pmatrix},\quad x:=\begin{pmatrix}x_1\\x_2\end{pmatrix} ,\quad \|x\|_2^2=1=x_1^2+x_2^2$$ and define $$ F(x)=x^T\cdot A\cdot x=(x_1, x_2)\cdot \begin{pmatrix}a&b\\b&c\end{pmatrix} \cdot \begin{pmatrix}x_1\\x_2\end{pmatrix}\\=(x_1, \sqrt{1-x_1^2})\cdot \begin{pmatrix}a&b\\b&c\end{pmatrix} \cdot \begin{pmatrix}x_1\\\sqrt{1-x_1^2}\end{pmatrix}=:F(x_1) $$.

Derivative: $$ \nabla F(x_1)=\left(1,\frac{-x_1}{\sqrt{1-x_1^2}}\right)\cdot \begin{pmatrix}a&b\\b&c\end{pmatrix} \cdot \begin{pmatrix}1\\\frac{-x_1}{\sqrt{1-x_1^2}}\end{pmatrix}\stackrel{!}{=}0 $$.

From here I get stuck and I don't see how can I use the information about $ A $, the eigenvalues and the rayleigh qoutient.

hallo007
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1 Answers1

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Note that the set $\|x\|=1$ is compact, so a $\max$ exists.

Solve $\max_{ \|x\|^2 =1 } {1 \over 2}x^T A x$ using Lagrange multipliers to get $Ax = \lambda x$, so we see that the Lagrange multiplier must be an eigenvalue.

Hence $\max_{ \|x\|^2 =1 } x^T A x = \lambda_\max$ and in the given example, since the eigenvalues are distinct we see that $x$ must be an eigenvector and hence the two solutions are $\pm x$.

copper.hat
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  • Ok, I used the Lagrange multipliers and got $ Ax=\lambda x $ and because $ |x|_2=1 $ describes a compact set max does exist. So far so good. Because the eigenvalues are different we have at least two critical points. But how do you came to the two solutions $ \pm x $ for each eigenvalue? – hallo007 Jan 05 '21 at 19:40
  • Well, $\ker (A-\lambda_1 I)$ is a one dimensional subspace and this intersects the set ${ y | |y|=1 }$ at two places. – copper.hat Jan 05 '21 at 19:47
  • I uderstand. So we four critical points. How can I determine how many maximizers are existing? – hallo007 Jan 05 '21 at 19:52
  • Since $\lambda_1 > \lambda_2$ you know they are the feasible points corresponding to the eigenvalue $\lambda_1$. – copper.hat Jan 05 '21 at 19:55
  • I don't see it how I can determine it with the fact $ \lambda_1>\lambda_2 $. – hallo007 Jan 05 '21 at 20:14
  • The Lagrange multipliers restrict the possible solutions to eigenvectors. for each eigenvector the cost is given by its eigenvalue, so the $\max$ must occur with the largest eigenvalue. – copper.hat Jan 05 '21 at 20:27