Consider the system of equations described by

\(\begin{cases}x_1=2x_1-3x_2\\x_2=4x_1-5x_2\end{cases}\)

The system of equations in the matrix form

\(\left(\begin{array}{c}x_{1}\\ x_{2}\end{array}\right)=\begin{bmatrix}2 & -3 \\4 & -5 \end{bmatrix} \left(\begin{array}{c}x_{1}\\ x_{2}\end{array}\right)\)

\(\Rightarrow x'=Ax,\ where\ A=\begin{bmatrix}2 & -3 \\4 & -5 \end{bmatrix}, x=(x_{1}, x_{2})\)

Now, \(|A-\lambda I|=\begin{bmatrix}2-\lambda & -3 \\4& -5-\lambda \end{bmatrix}=0\)

\(\displaystyle\Rightarrow\lambda^{{{2}}}+{3}\lambda+{2}={0}\Rightarrow\lambda=-{2},-{1}\)

Hence, -1,-2 are the eigen values.

\(\displaystyle{I}{f},\lambda=-{1}\)

\(Then \begin{bmatrix}3 & -3 \\4 & -4 \end{bmatrix} \left(\begin{array}{c}x_{1}\\ x_{2}\end{array}\right)=\left(\begin{array}{c}0\\ 0\end{array}\right)\)

\(\displaystyle\Rightarrow{3}{x}_{{{1}}}-{3}{x}_{{{2}}}={0}\Rightarrow{x}_{{{1}}}={x}_{{{2}}}={t}\epsilon\mathbb{R}\)

Hence, \(V_{1}=\left(\begin{array}{c}1\\ 1\end{array}\right)\)

\(\displaystyle{I}{f},\lambda=-{2}\)

\({T}{h}{e}{n},\begin{bmatrix}4 & -3 \\4 & -3 \end{bmatrix}\left(\begin{array}{c}x_{1}\\ x_{2}\end{array}\right)=\left(\begin{array}{c}0\\ 0\end{array}\right)\)

\(\displaystyle\Rightarrow{4}{x}_{{{1}}}-{3}{x}_{{{2}}}={0}\)

\(\displaystyle\Rightarrow{x}_{{{1}}}={\frac{{{3}}}{{{4}}}}{x}_{{{2}}}={0}\)

Hence, \(V_{2}=\left(\begin{array}{c}1\\ \frac{4}{3}\end{array}\right)\)

Hence,\(V_{1}\ \text{and}\ V_{2}\) are the associated eigenvectors.