Given that $A=\begin{bmatrix} 1 &3 &2 \\ 0 &4 &2 \\ 0 &-3 &-1 \end{bmatrix}$
We know that $AX=\lambda X$----->$(1),$Where $X=\begin{bmatrix} x\\y \\z \end{bmatrix}$
$\Rightarrow AX-\lambda X=[0]$
$\Rightarrow (A-\lambda I) X=[0]$------->$(2),$Where $I=\begin{bmatrix} 1 &0 &0 \\ 0 &1 &0 \\ 0 &0 &1 \end{bmatrix}$
Now,we can find $A-\lambda I=\begin{bmatrix} 1 &3 &2 \\ 0 &4 &2 \\ 0 &-3 &-1 \end{bmatrix}-\lambda \begin{bmatrix} 1 &0 &0 \\ 0 &1 &0 \\ 0 &0 &1 \end{bmatrix}$
$\Rightarrow A-\lambda I=\begin{bmatrix} 1 &3 &2 \\ 0 &4 &2 \\ 0 &-3 &-1 \end{bmatrix}-\begin{bmatrix} \lambda &0 &0 \\ 0 &\lambda &0 \\ 0 &0 &\lambda \end{bmatrix}$
$\Rightarrow A-\lambda I=\begin{bmatrix} 1-\lambda &3 &2 \\ 0 &4-\lambda &2 \\ 0 &-3 &-1-\lambda \end{bmatrix}$
Now,put the value in equation $(2)$
$(A-\lambda I)X=\begin{bmatrix} 1-\lambda &3 &2 \\ 0 &4-\lambda &2 \\ 0 &-3 &-1-\lambda \end{bmatrix}.\begin{bmatrix} x\\y \\z \end{bmatrix}$
$\Rightarrow \begin{bmatrix} 1-\lambda &3 &2 \\ 0 &4-\lambda &2 \\ 0 &-3 &-1-\lambda \end{bmatrix}.\begin{bmatrix} x\\y \\z \end{bmatrix}=\begin{bmatrix} 0\\0 \\0 \end{bmatrix}$
$For$ Eigen value $\lambda=1,$we can find the eigen vector
$\begin{bmatrix} 1-1&3 &2 \\ 0 &4-1 &2 \\ 0 &-3 &-1-1 \end{bmatrix}.\begin{bmatrix} x\\y \\z \end{bmatrix}=\begin{bmatrix} 0\\0 \\0 \end{bmatrix}$
$\begin{bmatrix} 0&3 &2 \\ 0 &3&2 \\ 0 &-3 &-2 \end{bmatrix}.\begin{bmatrix} x\\y \\z \end{bmatrix}=\begin{bmatrix} 0\\0 \\0 \end{bmatrix}$
Now we can perform some operation into the matrix,which is not effect to the rank,
$R_{3}\rightarrow R_{3}+R_{2}$
$R_{2}\rightarrow R_{2}-R_{1}$
$\begin{bmatrix} 0&3 &2 \\ 0 &0&0 \\ 0 &0 &0 \end{bmatrix}.\begin{bmatrix} x\\y \\z \end{bmatrix}=\begin{bmatrix} 0\\0 \\0 \end{bmatrix}$
Rank of the matrix$=1$ and Number of variables(Unknowns$=3$)
$r(A)=1,UK=3$
$r(A)<UK$,So,we can assign $3-1=2$ values to the two variables and solve the equation and get the solutions.
So, Linearly independent Eigen vector$=3-1=2$
Now,i can extend a bit,so it will be usefull.
When $R(A)<UK$,So,this is the case of infinite many number of solutions.
and We can assign the $UK-r(A)=3-1=2$ different values to the different variable and solve the equations.
$\Rightarrow\begin{bmatrix} 0&3 &2 \\ 0 &0&0 \\ 0 &0 &0 \end{bmatrix}.\begin{bmatrix} x\\y \\z \end{bmatrix}=\begin{bmatrix} 0\\0 \\0 \end{bmatrix}$
Let we can assign $x=k_{1}$ and $y=k_{2}$
Now $,0.x+3y+2z=0$
Here$,0.k_{1}+3.K_{2}+2z=0$
$\Rightarrow 0+3k_{2}+2z=0$
$\Rightarrow 3k_{2}+2z=0$
$\Rightarrow 2z=-3k_{2}$
$\Rightarrow z=-\frac{3}{2}k_{2}$
So,Eigen Vector $X=\begin{bmatrix} x\\y \\z \end{bmatrix}=\begin{bmatrix} k_{1}\\k_{2} \\-\frac{3}{2}k_{2} \end{bmatrix} =$ $k_1 \begin{bmatrix} 1\\ 0\\ 0 \end{bmatrix} + k_2 \begin{bmatrix} 0\\ 1\\ \frac{-3}{2} \end{bmatrix}$
You can clearly see$,x,y$ are independent and $z$ are dependent on $y$
For Finding the all Eigen Values,we can do something better,
$|A-\lambda I|=0$
$\Rightarrow\begin{vmatrix} 1-\lambda &3 &2 \\ 0 &4-\lambda &2 \\ 0 &-3 &-1-\lambda \end{vmatrix}=0$
$\Rightarrow (1-\lambda)\begin{vmatrix} 4-\lambda& 2\\-3 &-1-\lambda \end{vmatrix}=0$
$\Rightarrow (1-\lambda)[(4-\lambda)(-1-\lambda)-(-3)(2)]=0$
$\Rightarrow (1-\lambda)[(4-\lambda)(-1-\lambda)+6]=0$
$\Rightarrow (1-\lambda)[-4-4\lambda+\lambda+\lambda^{2}+6]=0$
$\Rightarrow (1-\lambda)[\lambda^{2}-3\lambda+2]=0$
$\Rightarrow (1-\lambda)[\lambda^{2}-2\lambda-\lambda+2]=0$
$\Rightarrow (1-\lambda)[\lambda(\lambda-2)-1(\lambda-2)]=0$
$\Rightarrow (1-\lambda)[(\lambda-2)(\lambda-1)]=0$
$\Rightarrow (1-\lambda)(\lambda-1)(\lambda-2)=0$
$\Rightarrow (-1) (\lambda-1)(\lambda-1)(\lambda-2)=0$
$\Rightarrow (\lambda-1)(\lambda-1)(\lambda-2)=0$
So$,\lambda=1,1,2$
We can also write like this $\lambda_{1}=1,\lambda_{2}=1,\lambda_{3}=2$
Another Method to find the Eigen Values,
$A=\begin{bmatrix} 1 &3 &2 \\ 0 &4 &2 \\ 0 &-3 &-1 \end{bmatrix}$
Let Eigen Vlaues are $\lambda_{1},\lambda_{2},\lambda_{3}$ Because we want to find the Eigen values of $3\times3$ matrix.
Important Properties for finding the Eigen Values:
$(1)$ Sum of All Eigen Values$=$Trace of the Matrix(Sum of all Leading Diagonal Elements)
$(2)$ Product of All Eigen Values$=Det(A)=|A|$
Here,$\lambda_{1}+\lambda_{2}+\lambda_{3}=1+4-1$
$\Rightarrow \lambda_{1}+\lambda_{2}+\lambda_{3}=4$-------------->$(3)$
and $\lambda_{1}.\lambda_{2}.\lambda_{3}=|A|$
$\lambda_{1}.\lambda_{2}.\lambda_{3}=1(-4+6)-0+0$
$\Rightarrow \lambda_{1}.\lambda_{2}.\lambda_{3}=2$----------------->$(4)$