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Eigenvector:An eigenvector ofAis a nonzero vector Vin${\mathbf{R}}_{\mathbf{n}}$such that$\mathbf{Av}\mathbf{=}\mathbf{位惫}$, for some scalar' $\mathit{位}$.

Consider the matrix$A=\left(\begin{array}{cc}2& 0\\ 3& 4\end{array}\right)$.

The objective is to show that 2 and 4 are eigen values of A and then to determine the corresponding eigenvectors.

Further, the objective is to determine the eigen basis for tA and then diagonalize x A

To determine the eigen values of A, solve the characteristic equation $\left|A-位I\right|=0$.

$$\begin{gathered} \left|A-位I\right|=0 \\ \left|\begin{array}{cc}2-位& 0\\ 3& 4-位\end{array}\right|=0 \\ \left(2-位\right)\left(4-位\right)=0 \\ 位=2,位=4 \end{gathered}$$

This proves that 2 and 4 are eigen values of matrix .

Recall the fact that if$位$is an eigen value of matrix A.

Then, it satisfies$A\stackrel{鈬赌}{v}=位\stackrel{鈬赌}{v}$.

Since 2 is an eigen value of A with eigenvector$\stackrel{鈬赌}{v}$. This implies as follows:

$$\begin{gathered} A\stackrel{鈬赌}{v}=2\stackrel{鈬赌}{v}\left(\begin{array}{cc}2& 0\\ 3& 4\end{array}\right)\left[\begin{array}{c}{v}_1\\ v_2\end{array}\right] \\ =2\left[\begin{array}{c}{v}_1\\ v_2\end{array}\right]\left[\begin{array}{c}2v_1\\ 3v+4v_2\end{array}\right] \\ =2\left[\begin{array}{c}{v}_1\\ v_2\end{array}\right] \end{gathered}$$

The solution to the above system of equations is,

$$\begin{gathered} 2v_1 \\ =2v_{1}3v_1+4v_1 \\ =2v_{2}3v_1 \\ =-2v_2 \end{gathered}$$

Or equivalently,

$\stackrel{鈬赌}{v}=t\left[\begin{array}{c}-2\\ 3\end{array}\right],t鈭�\mathrm{鈩�}$

Thus,data-custom-editor="chemistry" $\left[\begin{array}{c}-2\\ 3\end{array}\right]$is an eigenvector of A corresponding to eigen value $位=2$.

Since 4 is an eigen value of A with eigenvector$\stackrel{鈬赌}{v}$.

This implies as follows:

$$\begin{gathered} A\stackrel{鈬赌}{v}=4\stackrel{鈬赌}{v} \\ \left(\begin{array}{cc}2& 0\\ 3& 4\end{array}\right)\left[\begin{array}{c}{v}_1\\ v_2\end{array}\right]=4\left[\begin{array}{c}{v}_1\\ v_2\end{array}\right]\left[\begin{array}{c}2v_1\\ 3v_1+4v_2\end{array}\right] \\ =4\left[\begin{array}{c}{v}_1\\ v_2\end{array}\right] \end{gathered}$$

The solution to the above system of equations is,

$$\begin{gathered} 2v_1=4v_{1}3v_1+4v_2 \\ =4v_{2}3v_1 \\ =0 \end{gathered}$$

This implies,

$$\begin{gathered} v_1=0 \\ {v}_2鈭�\mathrm{鈩�} \end{gathered}$$

Or equivalently

$\stackrel{鈬赌}{v}=t\left[\begin{array}{c}-2\\ 3\end{array}\right]$

Thus,$\left[\begin{array}{c}-2\\ 3\end{array}\right]$ is an eigenvector of A corresponding to eigen value$位=2$

Thus, the vectors${\stackrel{鈬赌}{v}}_1=\left[\begin{array}{c}-2\\ 3\end{array}\right]\mathrm{and}{\stackrel{鈬赌}{\mathrm{v}}}_2=\left[\begin{array}{c}0\\ 1\end{array}\right]$andform an Eigen basis for A.

Since the eigenvalues and the Eigen basis are known, A is similar to the matrix B.

$$\begin{gathered} B=\left[\begin{array}{cc}{位}_1& 0\\ 0& {位}_2\end{array}\right] \\ =\left[\begin{array}{cc}2& 0\\ 0& 4\end{array}\right] \end{gathered}$$

The matrix that diagonalizes A is;

$$\begin{gathered} S=\left[\begin{array}{cc}{\stackrel{鈬赌}{v}}_1,& {\stackrel{鈬赌}{v}}_2\end{array}\right] \\ =\left[\begin{array}{cc}-2& 0\\ 3& 1\end{array}\right] \end{gathered}$$

Thus, $S=\left[\begin{array}{cc}-2& 0\\ 3& 1\end{array}\right]\begin{array}{}\end{array}$is a matrix that diagonalizes A.