91Ó°ÊÓ

$A=\left[\begin{array}{cc}8& -2\\ -2& 5\end{array}\right]$

Consider the matrix below:

$A=\left[\begin{array}{cc}8& -2\\ -2& 5\end{array}\right]$

Now, as shown below, calculate the eigen values:

$\det \left(\mathrm{A}-\mathrm{λ\pm ô}\right)=0$

$⇒\left|\begin{array}{cc}8-\mathrm{λ}& -2\\ -2& 5-\mathrm{λ}\end{array}\right|=0$

$$\begin{gathered} ⇒{\mathrm{λ}}^2-13\mathrm{λ}+36=0 \\ ⇒\left(\mathrm{λ}-4\right)·\left(\mathrm{λ}-9\right)=0 \end{gathered}$$

As a result, the eigenvalues are 9 and 4, yielding

$D=\left[\begin{array}{cc}9& 0\\ 0& 4\end{array}\right]$

So,

${\mathrm{D}}_1=\left[\begin{array}{cc}3& 0\\ 0& 2\end{array}\right]$

The eigen vectors are as follows:

$\left[\left.\begin{array}{c}-2\\ 1\end{array}\right]\mathrm{and}\left[\begin{array}{c}1\\ 2\end{array}\right.\right]$

Find S by combining the eigen vectors after they have been converted to unit vectors, as shown below.

$\frac{1}{\sqrt{5}}\left[\begin{array}{c}-2\\ 1\end{array}\right]\mathrm{and}\frac{1}{\sqrt{5}}\left[\begin{array}{c}1\\ 2\end{array}\right]$

As a result, we get S, as shown below:

$S=\frac{1}{\sqrt{5}}\left[\begin{array}{cc}-2& 1\\ 1& 2\end{array}\right]$

Now, as shown below, evaluate B:

$\mathrm{B}={\mathrm{SD}}_1\mathrm{S}$

$=\frac{1}{\sqrt{5}}\left[\begin{array}{cc}-2& 1\\ 1& 2\end{array}\right]\left[\begin{array}{cc}3& 0\\ 0& 2\end{array}\right]\frac{1}{\sqrt{5}}\left[\begin{array}{cc}-2& 1\\ 1& 2\end{array}\right]$

$=\frac{1}{\sqrt{5}}\left[\begin{array}{cc}-2& 1\\ 1& 2\end{array}\right]\left[\begin{array}{cc}3& 0\\ 0& 2\end{array}\right]\left[\begin{array}{cc}-2& 1\\ 1& 2\end{array}\right]$

$=\frac{1}{5}\left[\begin{array}{cc}-2& 1\\ 1& 2\end{array}\right]\left[\begin{array}{cc}-6& 3\\ 2& 4\end{array}\right]$

$⇒B=\frac{1}{5}\left[\begin{array}{cc}14& -2\\ -2& 11\end{array}\right]$

As a result, we can write A as:

$A=B^2$

$\mathrm{A}={\mathrm{B}}^2\mathrm{where},\mathrm{B}=\frac{1}{5}\left[\begin{array}{cc}14& -2\\ -2& 11\end{array}\right]$