 # I am given the following problem: A = [ <mtable rowspacing="4pt" columnspaci Sattelhofsk 2022-06-26 Answered
I am given the following problem:
$A=\left[\begin{array}{ccc}4& 1& 1\\ 1& 2& 3\\ 1& 3& 2\end{array}\right]$
Find
$\underset{x}{max}\frac{|\left(Ax,x\right)|}{\left(x,x\right)}$
where $\left(.,.\right)$ 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:
$E\left({\lambda }_{1}\right)=\text{span}\left({\left[\begin{array}{ccc}-2& 1& 1\end{array}\right]}^{T}\right)$
for ${\lambda }_{1}=3$ and
$E\left({\lambda }_{2}\right)=\text{span}\left({\left[\begin{array}{ccc}0& -1& 1\end{array}\right]}^{T}\right)$
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?
You can still ask an expert for help

• Live experts 24/7
• Questions are typically answered in as fast as 30 minutes
• Personalized clear answers

Solve your problem for the price of one coffee

• Math expert for every subject
• Pay only if we can solve it Xzavier Shelton
You wrote "for ${\lambda }_{3}=6$ there are no eigenvectors". This is not true !
We have, since $A$ is symmetric, that $\underset{x}{max}\frac{|\left(Ax,x\right)|}{\left(x,x\right)}=max\left\{\lambda :\lambda \in \sigma \left(A\right)\right\}$, where $\sigma \left(A\right)$ denotes the set of eigenvalues of $A$.
Hence
$\underset{x}{max}\frac{|\left(Ax,x\right)|}{\left(x,x\right)}=6.$

We have step-by-step solutions for your answer!