Consider the matrix

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

(c) For the values of k found in (b)

i. Find a non-zero vector vector `v_k` in the null space of `A-kI`

The other parts of the problem are here :

http://www.enotes.com/homework-help/consider-matrix-2-0-0-1-1-2-1-1-0-find-430769

http://www.enotes.com/homework-help/consider-matrix-2-0-0-1-1-2-1-1-0-b-which-values-430771

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Solving the characteristic equation

`det(A - kI) = 0`

We have

`(2-k)(k^2-k-2) = 0`

giving that the eigenvalues are `k=-1` or `2`.

A non-zero vector `v_k` in the null space of `A- kI` is the eigenvector of `A` paired with the eigenvalue `k`.

1) `k=-1`

To find the associated eigenvector we solve

`Av_k = kv_k`

`implies` `(A-kI)v_k = 0`

`implies` `([3,0,0],[1,2,-2],[1,-1,1])([x],[y],[z]) = 0`

` `Reducing to echelon form we get

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

`[[1,0,0],[0,1,-1],[0,-1,1]][[0],[0],[0]]` `implies` `[[1,0,0],[0,1,-1],[0,0,0]][[0],[0],[0]]`

Since this implies `x=0` and `y=z`, a particular eigenvector associated with the eigenvalue `k=-1` is

`v_1 = (0,1,1)`

2) Similarly for `k=2` we solve

`([0,0,0],[1,-1,-2],[1,-1,-2])([x],[y],[z]) = 0`

Immediately we see that rows 2 and 3 are identical. Letting `x=1` this implies that `y + 2z = 1`. Choosing the simple solution `y=1` and `z=0` we find that a particular eigenvector associated with the eigenvalue `k=2` is

`v_2 = (1,1,0)`

**The eigenvalues of A are -1 and 2. Vectors that are in the null** **space of A - kI are eigenvectors of A associated with an eigenvalue k.**

**An eigenvector associated with the eigenvalue k = -1 is v1 = (0,1,1). An eigenvector associated with the eigenvalue k = 2 is v2 = (1,1,0).**

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