# Find the Basis of an Eigenspace

## Homework Statement

Matrix A is 3x3
1 -5 -5
-5 1 5
5 -5 -9
Find basis for corresponding eigenspace when eigenvalue is -4

a) 0 b) 1 0 c) 1 d)1 0
1 0 1 0 0 1
-1 1 , -1 -1 -1 , 1

(A-lambda I)x=0

## The Attempt at a Solution

(A-(-4I)= 5 -5 -5
-5 5 5
5 -5 -5
that matrix times x will be equal to zero
created augmented matrix: 5 -5 -5 0
-5 5 5 0
5 -5 -5 0
find reduced echelon form of augmented matrix:
1 -1 -1 0
0 0 0 0
0 0 0 0
therefore x1=x2+x3
and x2 and x3 are free
vector x= x2+x3
x2
x3
which can be reduced to:
x2*1 + x3* 1
1 0
0 1

For the basis of the eigenspace, I then get:
1 1
1 0
0 , 1
However, the homework question is multiple choice and this is not one of the options. What am I doing wrong?

## Answers and Replies

Your notation is a bit hard to deciphere... What I can make up from it is that answer b consists of the basis vectors (in a transposed notation):
(1 0 1) and (0 1 -1)

while your answer is the basis:
(1 1 0) and (1 0 1)
But you can check that these two basis span the same subspace. In other words, your answer is some linear combination of the basis of answer b, namely:
(0 1 -1) = (1 1 0) - (1 0 1)

Thanks.
The notation did get kind of scrambled when I posted it, but thanks for your help!!