TITLE: What is the Maximum Value of $x^T Ax$?
QUESTION [7 upvotes]: Let $A$ be the matrix $\begin{bmatrix}3 &1 \\  1&2\end{bmatrix}$. What is the maximum value of $x^T Ax$ where the maximum is taken over all $x$ that are the unit eigenvectors of $A$?

$5$
$\frac{(5 + \sqrt{5})}{2}$
$3$
$\frac{(5 - \sqrt{5})}{2}$


Eigenvalues of $A$ are $\frac{(5 \pm \sqrt{5})}{2}$

My question is: what is $x^T Ax$ ? Can you explain little bit please?

REPLY [2 votes]: Here $A=\begin{bmatrix} 3 &1 \\ 1 & 2 \end{bmatrix}_{2\times 2}$
For finding the eigen value:
Chracteristic equation:$|A-\lambda I|=0$ where $I$ is called identity matrix.
                                     $\Rightarrow|A-\lambda I|=\begin{vmatrix} 3 &1 \\ 1 & 2 \end{vmatrix}-\lambda\begin{vmatrix}  1&0 \\ 0 & 1 \end{vmatrix}=0$
                                     $\Rightarrow \begin{vmatrix} 3 &1 \\ 1 & 2 \end{vmatrix}-\begin{vmatrix}  \lambda&0 \\ 0 & \lambda \end{vmatrix}=0$
                                     $\Rightarrow \begin{vmatrix} 3-\lambda &1 \\ 1 & 2-\lambda \end{vmatrix}=0$
                                     $\Rightarrow(3-\lambda)(2-\lambda)-1=0$
                                      $\Rightarrow 6-3\lambda-2\lambda+\lambda^{2}-1=0$
                                       $\Rightarrow \lambda^{2}-5\lambda+5=0$
Now$,\lambda =\frac{-(-5)\pm \sqrt{25-20}}{2}$
 $\lambda =\frac{5\pm \sqrt{5}}{2}$
 So$,\lambda_{1} =\frac{5+\sqrt{5}}{2}$ and $\lambda_{2} =\frac{5-\sqrt{5}}{2}$
Let Eigen vector $X=\begin{bmatrix} x_{1}\\x_{2} \end{bmatrix}_{2\times 1}$
Now find the Eigen Vectors:
           $AX=\lambda X$
       $\Rightarrow AX-\lambda X=\begin{bmatrix} 0 \end{bmatrix}$
       $\Rightarrow(A-\lambda I )X=\begin{bmatrix} 0 \end{bmatrix}$
  $\Rightarrow \begin{bmatrix} 3-\lambda &1 \\ 1 & 2-\lambda \end{bmatrix}\begin{bmatrix} x_{1}\\x_{2} \end{bmatrix}=\begin{bmatrix} 0\\0 \end{bmatrix}$------------$>(1)$
           Put $\lambda=\frac{5+\sqrt{5}}{2}$
  $\Rightarrow \begin{bmatrix} 3-(\frac{5+\sqrt{5}}{2}) &1 \\ 1 & 2-(\frac{5+\sqrt{5}}{2}) \end{bmatrix}\begin{bmatrix} x_{1}\\x_{2} \end{bmatrix}=\begin{bmatrix} 0\\0 \end{bmatrix}$
  $\Rightarrow \begin{bmatrix}\frac{6-5-\sqrt{5}}{2} &1 \\ 1 &\frac{4-5-\sqrt{5}}{2} \end{bmatrix}\begin{bmatrix} x_{1}\\x_{2} \end{bmatrix}=\begin{bmatrix} 0\\0 \end{bmatrix}$
  $\Rightarrow \begin{bmatrix}\frac{1-\sqrt{5}}{2} &1 \\ 1 &\frac{-1-\sqrt{5}}{2} \end{bmatrix}\begin{bmatrix} x_{1}\\x_{2} \end{bmatrix}=\begin{bmatrix} 0\\0 \end{bmatrix}$
Perform some row operation
$R_{1}\rightarrow\frac{1+\sqrt{5}}{2}R_{1}$
$\Rightarrow \begin{bmatrix}(\frac{1-\sqrt{5}}{2}).(\frac{1+\sqrt{5}}{2})&\frac{1+\sqrt{5}}{2} \\ 1 &\frac{-1-\sqrt{5}}{2} \end{bmatrix}\begin{bmatrix} x_{1}\\x_{2} \end{bmatrix}=\begin{bmatrix} 0\\0 \end{bmatrix}$
$\Rightarrow \begin{bmatrix}\frac{1-5}{4}&\frac{1+\sqrt{5}}{2} \\ 1 &\frac{-1-\sqrt{5}}{2} \end{bmatrix}\begin{bmatrix} x_{1}\\x_{2} \end{bmatrix}=\begin{bmatrix} 0\\0 \end{bmatrix}$
$\Rightarrow \begin{bmatrix}-1&\frac{1+\sqrt{5}}{2} \\ 1 &\frac{-1-\sqrt{5}}{2} \end{bmatrix}\begin{bmatrix} x_{1}\\x_{2} \end{bmatrix}=\begin{bmatrix} 0\\0 \end{bmatrix}$
Again do some operation $R_{2}\rightarrow R_{2}+R_{1}$
$\Rightarrow \begin{bmatrix}-1&\frac{1+\sqrt{5}}{2} \\ 0 &0 \end{bmatrix}\begin{bmatrix} x_{1}\\x_{2} \end{bmatrix}=\begin{bmatrix} 0\\0 \end{bmatrix}$
Here $Rank (A)=1$ and $number$  $of$  $unknowns$ $ =2$
clearly see $r(A)<UK(1<2),$So $UK-r(A)=2-1=1$ value assign to the one unknowns.[This is case of infinite solutions]
So$,x_{2}=k$ and $-x_{1}+(\frac{1+\sqrt{5}}{2})x_{2}=0$
                                  $ -x_{1}=-(\frac{1+\sqrt{5}}{2})k$
                                   $ x_{1}=(\frac{1+\sqrt{5}}{2})k$
Therefor $X=\begin{bmatrix} x_{1}\\x_{2} \end{bmatrix}=\begin{bmatrix}(\frac{1+\sqrt{5}}{2})k \\k \end{bmatrix}$
               $X=\begin{bmatrix} x_{1}\\x_{2} \end{bmatrix}=k \begin{bmatrix}(\frac{1+\sqrt{5}}{2}) \\1 \end{bmatrix}$
Now, find the unit vectors of $A:$
$\hat{u}=\frac{\vec{X}}{|\vec{X}|}$
$|\vec{X}|=\sqrt{(\frac{1+\sqrt{5}}{2})^{2}+1^{2}}=1.9$
$\hat{u}=\frac{\begin{bmatrix}(\frac{1+\sqrt{5}}{2}) \\1 \end{bmatrix}}{1.9}$
$\hat{u}=\begin{bmatrix} 0.85\\ 0.52 \end{bmatrix}$
here in given question  $\hat{u}=x=\begin{bmatrix} 0.85 \\0.52 \end{bmatrix}$ and $x^{T}=\begin{bmatrix} 0.85 &0.52 \end{bmatrix}$
Now $x^{T}Ax=\begin{bmatrix} 0.85 &0.52 \end{bmatrix}_{1\times 2}\times \begin{bmatrix} 3 &1 \\ 1 & 2 \end{bmatrix}_{2\times 2}\times \begin{bmatrix} 0.85 \\0.52 \end{bmatrix}_{2\times 1}$
$x^{T}Ax=\begin{bmatrix} 3.6 \end{bmatrix}_{1\times 1}$
and similarly do for $\lambda=\frac{5-\sqrt{5}}{2}$
 $X=\begin{bmatrix} x_{1}\\x_{2} \end{bmatrix}=k \begin{bmatrix}(\frac{1-\sqrt{5}}{2}) \\1 \end{bmatrix}$
Now, find the unit vectors of $A:$
$\hat{u}=\frac{\vec{X}}{|\vec{X}|}$
$|\vec{X}|=\sqrt{(\frac{1-\sqrt{5}}{2})^{2}+1^{2}}=1.2$
$\hat{u}=\frac{\begin{bmatrix}(\frac{1-\sqrt{5}}{2}) \\1 \end{bmatrix}}{1.2}$
$\hat{u}=\begin{bmatrix} -0.52\\ 0.83 \end{bmatrix}$
here in given question  $\hat{u}=x=\begin{bmatrix} -0.52\\0.83 \end{bmatrix}$ and $x^{T}=\begin{bmatrix} -0.52 &0.83 \end{bmatrix}$
Now $x^{T}Ax=\begin{bmatrix} -0.52 &0.83 \end{bmatrix}_{1\times 2}\times \begin{bmatrix} 3 &1 \\ 1 & 2 \end{bmatrix}_{2\times 2}\times \begin{bmatrix} -0.52 \\0.83 \end{bmatrix}_{2\times 1}$
$x^{T}Ax=\begin{bmatrix} 1.33 \end{bmatrix}_{1\times 1}$
So,I got maximum value $x^{T}Ax=\begin{bmatrix} 3.6 \end{bmatrix}_{1\times 1}=\begin{bmatrix} \frac{5+\sqrt{5}}{2} \end{bmatrix}$
and minimum value $x^{T}Ax=\begin{bmatrix} 1.33 \end{bmatrix}_{1\times 1}=\begin{bmatrix} \frac{5-\sqrt{5}}{2} \end{bmatrix}$