Answer:
Consider lambda as $
$ =0, & $ =6 (eigen values)
eigen vectors :
[1 2]
[0 0]
Step-by-step explanation:
1st step:
Lambda I
here I is identity matrix.
As our matrix is 2X2 , so we will take I= [ 1 0, 0 1]
consider lambda is represented as $
so $I matrix will be [$ 0, 0 $]
2nd Step:
A - Lambda I
i.e .[2 -2, -4 4] - [$ 0, 0 $]
=> [2-$ -2, -4 4-$]
3rd Step:
Det [A-$I]
i.e. [(2-$)(4-$) - (-2)(-4)] = 0
=>8-2$-4$+$²-8
=>$²-6$
4th Step:
Det [A-$I]=0
i.e. $²-6$ =0
$($-6)=0
i.e. $ =0, & $ =6 (eigen values)
5th Step
put these eigen values in [A-$I] matrix
i.e if we put $ =0
we get [2 -2, -4 4]
consider this matrix as B
then
B X⁻= 0⁻ (⁻ is a bar sign notation)
[2 -2, -4 4] [x₁ x₂] =[0 0]
by row reduction
-2R₁+ R₂ -> R₂
so
2X₁-2X₂=0
X₁=X₂
ie eigen vector will be [0 0]
now consider
i.e if we put $ =6
we get [-4 -2, -4 -2]
consider this matrix as B
then
B X⁻= 0⁻ (⁻ is a bar sign notation)
[-4 -2, -4 -2] [x₁ x₂] =[0 0]
by row reduction
R₁- R₂ -> R₂
so
-4X₁-2X₂=0
2X₁=X₂
if we put X₁=1
x₂=2
So eigen vector will be
[1 2]
