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. The other parts of the problem are here :...

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rakesh05 | High School Teacher | (Level 1) Assistant Educator

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Given matrix is `A=[[2,0,0],[1,1,-2],[1,-1,0]]` .

now `det(A-kI_3)=-(k-2)^2(k+1)`  (As per the previous part of the question).

Now we want to find the non zero vector `v_k`  in the null space of A-kI.

Now,   `det(A-kI_3)=0` `rArr(k-2)^2(k+1)=0`

                                         or,   k=2,-1.

Now (i) for k=2, the matrix A-kI becomes

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

Now let `v_k=(x_1,x_2,x_3)` be a non zero vector. Then `v_k`  will be i the null space of A-kI if 

                     (A-kI)`v_k=0`

i.e.   `[[0,0,0],[1,-1,-2],[1,-1,-2]][[x_1],[x_2],[x_3]]=[[0],[0],[0]]`

i.e.    `x_1-x_2-2x_3=0`

`` and  `x_1-x_2-2x_3=0`

i.e.   `x_1-x_2-2x_3=0` . Here we see two free variables.

If we take `x_2`  and `x_3`  to be the free variables,

    `x_1=x_2+2x_3` .

Take `x_2=1, x_3=0` , we get `x_1=1.`  So the vector (1,1,0) is in the null space of A-kI.

if we take `x_2=0,x_3=1,`  we get `x_1=2.` So vector (2,0,1) will be in the null space of A-kI.

So both the vectors (1,1,0) and (2,0,1) will act for `v_k.`

 (ii) For k=-1. the matrix A-kI becomes `[[2,0,0],[1,1,-2],[1,-1,0]]-[[-1,0,0],[0,-1,0],[0,0,-1]]=[[3,0,0],[1,2,-2],[1,-1,1]]`

i.e. A+I=`[[3,0,0],[1,2,-2],[1,-1,1]]` .

Now the vector `v_k=(x_1,x_2,x_3) `  will be in the null space of A+I  if       `[[3,0,0],[1,2,-2],[1,-1,1]][[x_1],[x_2],[x_3]]=[[0],[0],[0]]`

`or` ,    `3x_1=0` ,  `x_1+2x_2-2x_3=0,` and `x_1-x_2+x_3=0`

or,     `x_1=0`  and `x_2=x_3` .

If we take `x_3=t` , then `v_k=t[[0],[1],[1]]` ,

so  `v_k=[[0],[1],[1]]`  is the required non zero vector in the null space of A-kI for k=-1.

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