Give that the matrix $A=\begin{bmatrix} 3 &-2 &2 \\ 0 &-2 &1 \\ 0&0 &1 \end{bmatrix}$

Let suppose the Eigen Vector $X=\begin{bmatrix} x\\ y\\z \end{bmatrix}$

We know that $AX=\lambda X$

$AX-\lambda X=\begin{bmatrix} 0 \end{bmatrix}$

$(A-\lambda I )X=\begin{bmatrix} 0 \end{bmatrix}$---->(1) where $I=\begin{bmatrix} 1 &0 &0 \\ 0&1 &0 \\ 0& 0 &1 \end{bmatrix}$Identity matrix.

$(A-\lambda I )=\begin{bmatrix} 3 &-2 &2 \\ 0&-2 &1 \\ 0& 0 &1 \end{bmatrix}-\lambda\begin{bmatrix} 1 &0 &0 \\ 0&1 &0 \\ 0& 0 &1 \end{bmatrix}$

Given that $\lambda=-2$

$(A-\lambda I )=\begin{bmatrix} 3 &-2 &2 \\ 0&-2 &1 \\ 0& 0 &1 \end{bmatrix}-(-2)\begin{bmatrix} 1 &0 &0 \\ 0&1 &0 \\ 0& 0 &1 \end{bmatrix}$

$(A-\lambda I )=\begin{bmatrix} 3 &-2 &2 \\ 0&-2 &1 \\ 0& 0 &1 \end{bmatrix}+2\begin{bmatrix} 1 &0 &0 \\ 0&1 &0 \\ 0& 0 &1 \end{bmatrix}$

$(A-\lambda I )=\begin{bmatrix} 3 &-2 &2 \\ 0&-2 &1 \\ 0& 0 &1 \end{bmatrix}+\begin{bmatrix} 2 &0 &0 \\ 0&2 &0 \\ 0& 0 &2 \end{bmatrix}$

$(A-\lambda I )=\begin{bmatrix} 5 &-2 &2 \\ 0&0 &1 \\ 0& 0 &3 \end{bmatrix}$

Now,from the equation$(1),$

$(A-\lambda I )X=\begin{bmatrix} 0 \end{bmatrix}$

$\begin{bmatrix} 5 &-2 &2 \\ 0&0 &1 \\ 0& 0 &3 \end{bmatrix}\begin{bmatrix} x\\y \\z \end{bmatrix}=\begin{bmatrix} 0\\0 \\0 \end{bmatrix}$

Now,we can apply some operation on matrix

$R_{2}\rightarrow3R_{2}$

$\begin{bmatrix} 5 &-2 &2 \\ 0&0 &3 \\ 0& 0 &3 \end{bmatrix}\begin{bmatrix} x\\y \\z \end{bmatrix}=\begin{bmatrix} 0\\0 \\0 \end{bmatrix}$

Again apply some operation on matrix

$R_{3}\rightarrow R_{3}-R_{2}$

$\begin{bmatrix} 5 &-2 &2 \\ 0&0 &3 \\ 0& 0 &0 \end{bmatrix}\begin{bmatrix} x\\y \\z \end{bmatrix}=\begin{bmatrix} 0\\0 \\0 \end{bmatrix}$

In $R_{3}$,we create Nullity.So Rank of this matrix is $2$ [Rank:Maximum number of independent Rows/Colums]

$ r=2,$ and number of unknowns ( Variables $=3$ )

Here clearly see, $r<n$

So,there are Infinitely many solution are possible,and $n-r=3-2=1 $variable give some value and other will be evaluate accordingly.

$\begin{bmatrix} 5x-2y+2z\\0x+0y+3z \\0x+0y+0z \end{bmatrix}=\begin{bmatrix} 0\\0 \\0 \end{bmatrix}$

Let say $y=k$

$z=0$ ---->(2)

$5x-2k+2z=0$

$5x-2k+0=0$

$5x=2k$

$x=\frac{2k}{5}$

Finally we got the Eigen Vector$,X=\begin{bmatrix} x\\y \\z \end{bmatrix}=\begin{bmatrix} \frac{2k}{5}\\ k \\ 0 \end{bmatrix}$

Put the Value of $k=5$

and we get$,X=\begin{bmatrix} x\\y \\z \end{bmatrix}=\begin{bmatrix} \frac{2*5}{5}\\ 5 \\ 0 \end{bmatrix}$

$X=\begin{bmatrix} x\\y \\z \end{bmatrix}=\begin{bmatrix} 2 \\ 5 \\ 0 \end{bmatrix}$

Actually,we get Infinitely many number of solutions.So,we can get Infinte number of Eigen Vector,but for $k=5$,we got one Eigen Vector.