# Thread: Find a basis for the eigenspace

1. ## Find a basis for the eigenspace

B = 3 -4 4 eigenvalue = -1

-2 1 -2

-4 4 -5

Find a basis for the eigenspace corresponding to the eigenvalue -1.

So far I have:
-1-3 -4 4
-2 -1-1 -2
-4 4 -1--5

-4 -4 4 (divde by row 2)
-2 -2 -2
-4 4 4 (divide by row 2)

2 2 -2
-2 -2 -2
2 -2 -2

I get here and this is where I get stuck.

2. Originally Posted by Arita
B = 3 -4 4 eigenvalue = -1

-2 1 -2

-4 4 -5

Find a basis for the eigenspace corresponding to the eigenvalue -1.

So far I have:
-1-3 -4 4
-2 -1-1 -2
-4 4 -1--5

-4 -4 4 (divde by row 2)
-2 -2 -2
-4 4 4 (divide by row 2)

2 2 -2
-2 -2 -2
2 -2 -2

I get here and this is where I get stuck.

Is B the eigenvector corresponding to the eigenvalue of -1? What was the original matrix?

3. B is the matrix.

The original matrix is:

3 -4 4
-2 1 -2
-4 4 -5

I then went to (-4, -4, 4) (-2, -2, -2)(-4, 4, 4) then reduced to (-1, -1, 1) (-2,-2,-2) (-4, 4, 4) to finally reducing to
1 -1 1
0 0 0
0 0 0

But how do I get the eigenvector

4. Originally Posted by Arita
B is the matrix.

The original matrix is:

3 -4 4
-2 1 -2
-4 4 -5
$\displaystyle\begin{vmatrix}3-\lambda&-4&4\\-2&1-\lambda&-2\\-4&4&-5-\lambda\end{vmatrix}=(3-\lambda)[-(1-\lambda)(-5-\lambda)+8]=-(\lambda+1)(\lambda^2-1)=0$

Eigenvalues

$\lambda_1=\lambda_2=-1, \ \lambda_3=1$

$\displaystyle\lambda_1=\lambda_2=-1, \ \ \begin{bmatrix}4&-4&4\\-2&2&-2\\-4&1&-4\end{bmatrix}\Rightarrow\mbox{rref}=\begin{bmatri x}1&-1&1\\0&0&0\\0&0&0\end{bmatrix}$

$\displaystyle x_1=x_2-x_3\Rightarrow\begin{bmatrix}x_2-x_3\\x_2\\x_3\end{bmatrix}\Rightarrow x_2\begin{bmatrix}1\\1\\0\end{bmatrix}+x_3\begin{b matrix}-1\\0\\1\end{bmatrix}$

5. Just so I understand this correctly, there are to eigenvectors for the eigenvalue of -1?

6. Originally Posted by Arita
Just so I understand this correctly, there are to eigenvectors for the eigenvalue of -1?
Yes, the eigenvectors for -1 are

$\displaystyle \left\{\begin{bmatrix}1\\1\\0\end{bmatrix},\begin{ bmatrix}-1\\0\\1\end{bmatrix}\right\}$

7. Thanks so much for the help!

8. Originally Posted by Arita
Thanks so much for the help!
There is also an eigenvector associate with lambda equals one too. Just plug it in to

$\displaystyle
\displaystyle\begin{vmatrix}3-\lambda&-4&4\\-2&1-\lambda&-2\\-4&4&-5-\lambda\end{vmatrix}$

and find the reduced row echelon form.