91Ó°ÊÓ

Eigenvector:An eigenvector ofAis a nonzero vector vin${\mathbf{R}}_{\mathbf{n}}$such that$\mathbf{Av}\mathbf{=}\mathbf{λ\pm ¹}$, for some scalar$\mathbf{λ}$.

If $\mathrm{λ}$is an Eigen value of A this means that:

$\left(A-\mathrm{λ}{I}_n\right)\stackrel{â‡\P Ä}{v}=\stackrel{â‡\P Ä}{0}$

This implies;

$A-\mathrm{λ}{I}_n$Fails to be invertible;

That is $det\left(A-\mathrm{λ}{I}_n\right)=0$

If $\stackrel{â‡\P Ä}{v}$is an Eigen vector of A this means that:

$A\stackrel{â‡\P Ä}{v}=\mathrm{λ}\text{'}v$

Now, let$A=\left[\begin{array}{cc}-6& 6\\ -15& 13\end{array}\right]$.

The objective is to show that $\mathrm{λ}=4$is an Eigen value of A.

To do this need to show that $A-\mathrm{λ}{I}_n$is NOT invertible for the given $\mathrm{λ}$.

$$\begin{gathered} A-4I_n \\ =\left[\begin{array}{cc}-6& 6\\ -15& 13\end{array}\right]-\left[\begin{array}{cc}4& 0\\ 0& 4\end{array}\right] \\ =\left[\begin{array}{cc}-10& 6\\ -15& 9\end{array}\right] \end{gathered}$$

This implies;

$$\begin{gathered} \det \left(\mathrm{A}-4I_2\right)=\det \left[\begin{array}{cc}-10& 6\\ -15& 9\end{array}\right] \\ =-90+90 \\ =0 \end{gathered}$$

Thus, 4 is an Eigen value of.

Hence, it is proved.

Now, find the Eigen vector for $\mathrm{λ}=4$and will have:

$$\begin{gathered} A\stackrel{â‡\P Ä}{v}=\mathrm{λ}\stackrel{â‡\P Ä}{v} \\ \left[\begin{array}{cc}-6& 6\\ -15& 13\end{array}\right]\stackrel{â‡\P Ä}{v}=4\stackrel{â‡\P Ä}{v} \end{gathered}$$

This implies;

$$\begin{gathered} \left[\begin{array}{cc}-6& 6\\ -15& 13\end{array}\right]\left[\begin{array}{c}{v}_1\\ v_2\end{array}\right]=4\left[\begin{array}{c}{v}_1\\ v_2\end{array}\right] \\ \left[\begin{array}{cc}-6v_1& 6v_2\\ -15v_1& 13v_2\end{array}\right]=\left[\begin{array}{c}4v_1\\ 4v_2\end{array}\right] \end{gathered}$$

This implies;

$$\begin{gathered} -6v_1+6v_2=4v_{2}10v_2 \\ =6v_{2}5v_1 \\ =3v_2 \end{gathered}$$

From here, it can be seen that all vectors of the form: $\left[\stackrel{â‡\P Ä}{v}=\left[\begin{array}{c}3t\\ 5t\end{array}\right]\right]$will be Eigen vectors of A .