1.31. Find the eigenvalues and eigenvectors of the linear operator $\hat{A}$ with the matrix

$$
A=\left(\begin{array}{ll}
1 & 2 \\
3 & 2
\end{array}\right)
$$

Will the eigenvectors be orthogonal?
Solution. Let's form the characteristic equation

$$
\left|\begin{array}{cc}
1-\lambda & 2 \\
3 & 2-\lambda
\end{array}\right|=0 \Rightarrow(1-\lambda)(2-\lambda)-6=0 \Rightarrow \lambda^{2}-3 \lambda-4=0
$$

Solving the quadratic equation, we find the eigenvalues $\lambda_{1}=-1, \lambda_{2}=4$. Next, we will find the eigenvector corresponding to the eigenvalue $\lambda_{1}=-1$. We set up the system

$$
\left\{\begin{array} { l } 
{ ( 1 - \lambda _ { 1 } ) x _ { 1 } + 2 x _ { 2 } = 0 , } \\
{ 3 x _ { 1 } + ( 2 - \lambda _ { 1 } ) x _ { 2 } = 0 . }
\end{array} \Rightarrow \left\{\begin{array}{l}
2 x_{1}+2 x_{2}=0, \\
3 x_{1}+3 x_{2}=0 .
\end{array} \Rightarrow x_{1}+x_{2}=0\right.\right.
$$

Let $x_{1}=m$, where $m$ is any non-zero real number. The eigenvector corresponding to the eigenvalue $\lambda_{1}$ is

$$
x^{(1)}=m \vec{i}-m \vec{j}=\binom{m}{-m}
$$

For the eigenvector $\lambda_{2}=4$, we similarly get

$$
\left\{\begin{array} { l } 
{ ( 1 - \lambda _ { 2 } ) x _ { 1 } + 2 x _ { 2 } = 0 , } \\
{ 3 x _ { 1 } + ( 2 - \lambda _ { 2 } ) x _ { 2 } = 0 . }
\end{array} \Rightarrow \left\{\begin{array}{l}
-3 x_{1}+2 x_{2}=0, \\
3 x_{1}-2 x_{2}=0 .
\end{array} \Rightarrow 3 x_{1}-2 x_{2}=0\right.\right.
$$

Let $x_{1}=m$, where $m$ is any non-zero real number, then

$$
x^{(2)}=m \vec{i}+1.5 m \vec{j}=\binom{m}{1.5 m}
$$

and the first eigenvector corresponding to the eigenvalue $\lambda_{1}$ is

$$
x^{(1)}=m \vec{i}-m \vec{j}=\binom{m}{-m}
$$

Let's find the dot product of the eigenvectors: $x^{(1)} \cdot x^{(2)}=m \cdot m+(-m) \cdot 1.5 m=-0.5 m^{2}$. For the vectors to be orthogonal, their dot product must be zero, but then $m=0$, which is not possible because by definition an eigenvector cannot be the zero vector. Therefore, the eigenvectors are not orthogonal.