Sattelhofsk

2022-06-26

I am given the following problem:

$A=\left[\begin{array}{ccc}4& 1& 1\\ 1& 2& 3\\ 1& 3& 2\end{array}\right]$

Find

$\underset{x}{max}\frac{|(Ax,x)|}{(x,x)}$

where $(.,.)$ is a dot product of vectors and the maximization is performed over all $x={\left[\begin{array}{ccc}{x}_{1}& {x}_{2}& {x}_{3}\end{array}\right]}^{T}\in {\mathbb{R}}^{3}$, such that $\sum _{i=1}^{3}{x}_{i}=0$

I have found the eigenvectors for $A$ and they happen to match the sum criterion:

$E({\lambda}_{1})=\text{span}({\left[\begin{array}{ccc}-2& 1& 1\end{array}\right]}^{T})$

for ${\lambda}_{1}=3$ and

$E({\lambda}_{2})=\text{span}({\left[\begin{array}{ccc}0& -1& 1\end{array}\right]}^{T})$

for ${\lambda}_{2}=-1$.

(For ${\lambda}_{3}=6$ there are no eigenvectors).

Can the above eigenvectors and eigenvalues be used for solving this maximization problem?

$A=\left[\begin{array}{ccc}4& 1& 1\\ 1& 2& 3\\ 1& 3& 2\end{array}\right]$

Find

$\underset{x}{max}\frac{|(Ax,x)|}{(x,x)}$

where $(.,.)$ is a dot product of vectors and the maximization is performed over all $x={\left[\begin{array}{ccc}{x}_{1}& {x}_{2}& {x}_{3}\end{array}\right]}^{T}\in {\mathbb{R}}^{3}$, such that $\sum _{i=1}^{3}{x}_{i}=0$

I have found the eigenvectors for $A$ and they happen to match the sum criterion:

$E({\lambda}_{1})=\text{span}({\left[\begin{array}{ccc}-2& 1& 1\end{array}\right]}^{T})$

for ${\lambda}_{1}=3$ and

$E({\lambda}_{2})=\text{span}({\left[\begin{array}{ccc}0& -1& 1\end{array}\right]}^{T})$

for ${\lambda}_{2}=-1$.

(For ${\lambda}_{3}=6$ there are no eigenvectors).

Can the above eigenvectors and eigenvalues be used for solving this maximization problem?

Xzavier Shelton

Beginner2022-06-27Added 26 answers

You wrote "for ${\lambda}_{3}=6$ there are no eigenvectors". This is not true !

We have, since $A$ is symmetric, that $\underset{x}{max}\frac{|(Ax,x)|}{(x,x)}=max\{\lambda :\lambda \in \sigma (A)\}$, where $\sigma (A)$ denotes the set of eigenvalues of $A$.

Hence

$\underset{x}{max}\frac{|(Ax,x)|}{(x,x)}=6.$

We have, since $A$ is symmetric, that $\underset{x}{max}\frac{|(Ax,x)|}{(x,x)}=max\{\lambda :\lambda \in \sigma (A)\}$, where $\sigma (A)$ denotes the set of eigenvalues of $A$.

Hence

$\underset{x}{max}\frac{|(Ax,x)|}{(x,x)}=6.$

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