We know that for Eigenvector $AX=\lambda X$------------------>$(1)$ Where A=Given Matrix and $X=\begin{bmatrix} x_{1}\\x_{2} \end{bmatrix}$ Eigen Vectors
$\Rightarrow AX-\lambda X=[0]$
$\Rightarrow (A-\lambda I) X=[0]$---------------->$(2)$Where $I=\begin{bmatrix} 1 & 0\\ 0 &1 \end{bmatrix}$Idenitity Matrix.
For Eigen Values we write the characteristic equation$:|A-\lambda I|=0$----------->$(3)$
Now,Given that Matrix $I=\begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix}$
First we can find the Eigen values:
Here in this question $A=I$
So,we can write the equation$:|I-\lambda I|=0$------->$(4)$
now we can find $(I-\lambda I)=\begin{bmatrix} 1 & 0\\ 0 &1 \end{bmatrix}-\lambda \begin{bmatrix} 1 & 0\\ 0 &1 \end{bmatrix}$
$(I-\lambda I)=\begin{bmatrix} 1 & 0\\ 0 &1 \end{bmatrix}- \begin{bmatrix} \lambda & 0\\ 0 &\lambda \end{bmatrix}$
$(I-\lambda I)=\begin{bmatrix} 1-\lambda & 0\\ 0 &1-\lambda \end{bmatrix}$
Now from the equation $(4),$
$|I-\lambda I|=\begin{vmatrix} 1-\lambda & 0\\ 0 &1-\lambda \end{vmatrix}=0$
$\Rightarrow [(1-\lambda)(1-\lambda)-0]=0$
So$,\lambda=1,1$
Eigen values are $\lambda_{1}=1,\lambda_{2}=1$
We have Another Method,to finding the Eigen Values:
Given that Matrix $I=\begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix}$
Let suppose Eigen Values are $\lambda_{1},\lambda_{2}$
Important Property of finding the Eigen Values:
$(1)$ Sum of All Eigen Values$=$Sum of All Leading Diagonal Elements(Trace of the Matrix)
$(2)$ Product of All Eigen Values $=Det(A)=|A|$
Now,$\lambda_{1}+\lambda_{2}=1+1=2$----------->$(5)$
and $\lambda_{1}.\lambda_{2}=|I|=(1-0)=1$------------->$(6)$
Now we can find the values of $\lambda_{1},\lambda_{2}$
We know that if $a,b$ are the root the quadratic equation,so we can construct the quadratic equation
$x^{2}-(a+b)x+ab=0$
If $\lambda_{1},\lambda_{2}$ are the root of the quadratic eqation we can write the quadratic equation
In our question we have the value of $\lambda_{1}+\lambda_{2}=2,\lambda_{1}.\lambda_{2}=1$
$x^{2}-(\lambda_{1}+\lambda_{2})x+\lambda_{1}.\lambda_{2}=0$
$x^{2}-2x+1=0$
$x^{2}+1^{2}-2x=0$
$(x-1)^{2}=0$
$x=1,1$
So$,\lambda_{1}=1,\lambda_{2}=1$
We have Another Method,to finding the Eigen Values:
Given that Matrix $I=\begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix}$
In the case of Triangular matrix(Lower triangular matrix or Upper triangular matrix),Leading diagonal elements itself are Eigen Values.
We all know that Identity Matrix is the triangular matrix $($ Symmetric matrix also $A^{T}=A$$)$
So,Eigen Values are $\lambda_{1}=1,\lambda_{2}=1$
Now, from the equation $(2)$
$(A-\lambda I) X=[0]$
Here $A=I,$ $(I-\lambda I) X=[0]$
$(I-\lambda I)X=\begin{bmatrix} 1-\lambda & 0\\ 0 &1-\lambda \end{bmatrix}.\begin{bmatrix} x_{1}\\x_{2} \end{bmatrix}=\begin{bmatrix} 0\\0 \end{bmatrix}$
Here $\lambda=1$
$\Rightarrow \begin{bmatrix} 1-1 & 0\\ 0 &1-1 \end{bmatrix}.\begin{bmatrix} x_{1}\\x_{2} \end{bmatrix}=\begin{bmatrix} 0\\0 \end{bmatrix}$
$\Rightarrow \begin{bmatrix}0 & 0\\ 0 &0\end{bmatrix}.\begin{bmatrix} x_{1}\\x_{2} \end{bmatrix}=\begin{bmatrix} 0\\0 \end{bmatrix}$
Rank of the matrix $,r(I)=0$ ,Number of unknowns $=2$
Here$,r(I)<UK (0<2),$ this is the condition for Infinite many numbers of solutions, and we can assign $UK−r(I)=2−0=2$ linearly independent values to different variable for finding the $x_{1},x_{2}$ $[$UK---->Number of Unknown$]$
Here $x_{1}=k_{1},x_{2}=k_{2}$
Now $,0.x_{1}+0.x_{2}=0$------------>$(7)$
Now,$X=\begin{bmatrix} x_{1}\\x_{2} \end{bmatrix}$
$\Rightarrow \begin{bmatrix} x_{1}\\x_{2} \end{bmatrix}=\begin{bmatrix} k_{1}\\k_{2} \end{bmatrix} = k_1 \begin{bmatrix} 1\\ 0 \end{bmatrix} + k_2 \begin{bmatrix} 0\\ 1 \end{bmatrix}$
You can clearly see,there are two linearly independent eigen vector