# Consider the matrix A = [[2, 0, 0],[1,1,-2],[1,-1, 0]] (c) For the values of k in (a) , i. Find a non-zero vector vector v_kin the null space of A -kI.

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From the previous parts we can get k = -1 and k = 2.

When k=-1;

`[[2-(-1), 0, 0],[1, 1-(-1),-2],[1,-1,-(-1)]]`

`= [[3, 0, 0],[1, 2,-2],[1,-1,1]]`

`= [[1, 0, 0],[1, 2,-2],[1,-1,1]]`

`= [[1, 0, 0],[0, 2,-2],[0,-1,1]]`

Vector v-1 = [0,1,1]^T is an eigenvector of -1

`Av_(-1)`

`= [[2, 0, 0],[1,1,-2],[1,-1, 0]] [0,1,1]^T `

`= [0, -1, -1]^T`

`= -v_(-1)`

When k=2

`[[2-(2), 0, 0],[1, 1-(2),-2],[1,-1,-(2)]]`

`= [[0, 0, 0],[1, -1,-2],[1,-1,-2]]`

`= [[0, 0, 0],[1, -1,-2],[0,0,0]]`

Vectors `v_2 =[1, 1, 0]^T` and `v_(2') = [2,0,1]^T` are both eigenvectors

`Av_2`

`= [[2, 0, 0],[1,1,-2],[1,-1, 0]] [1,1,0]^T `

`=[2, 2, 0]^T`

`= 2v_2`

`Av_(2')`

`= [[2, 0, 0],[1,1,-2],[1,-1, 0]] [2,0,1]^T `

`= [4, 0, 2]^T`

`= 2v_(2')`

*Hence it is proved that A(vector v_k)= k(vector v_k)*

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