Here $A=\begin{bmatrix} 3 &1 \\ 1 & 2 \end{bmatrix}_{2\times 2}$
For finding the eigen-value:
Characteristics 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 $A:$
$\hat{u}=\frac{\vec{X}}{|\vec{X}|}$
$|\vec{X}|=\sqrt{(\frac{1+\sqrt{5}}{2})^{2}+1^{2}}=1.9$
$\hat{u}=\frac{\begin{bmatrix}(\frac{1+\sqrt{5}}{2}) \\1 \end{bmatrix}}{1.9}$
$\hat{u}=\begin{bmatrix} 0.85\\ 0.52 \end{bmatrix}$ $[$For unit vector $|\hat{u}|=\sqrt{(0.85)^{2}+(0.52)^{2}}=1$ Always$.]$
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 $x^{T}Ax=\begin{bmatrix} 1.33 \end{bmatrix}_{1\times 1}=\begin{bmatrix} \frac{5-\sqrt{5}}{2} \end{bmatrix}$
A unit vector is a vector of length 1, sometimes also called a direction vector.The unit vector $\hat{v}$ having the same direction as a given (non-zero) vector $\vec{v}$ have.
Unit vector is defined as $\hat{v}=\frac{\vec{v}}{|\vec{v}|}$ and $|\hat{v}|=1$ $always.$