# Find a basis for the eigenspace

• January 14th 2011, 03:23 PM
Arita
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.

• January 14th 2011, 03:36 PM
dwsmith
Quote:

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?
• January 14th 2011, 03:43 PM
Arita
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
• January 14th 2011, 04:03 PM
dwsmith
Quote:

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

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

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

$\displaystyle x_1=x_2-x_3\Rightarrow\begin{bmatrix}x_2-x_3\\x_2\\x_3\end{bmatrix}\Rightarrow x_2\begin{bmatrix}1\\1\\0\end{bmatrix}+x_3\begin{b matrix}-1\\0\\1\end{bmatrix}$
• January 14th 2011, 04:30 PM
Arita
Just so I understand this correctly, there are to eigenvectors for the eigenvalue of -1?
• January 14th 2011, 05:33 PM
dwsmith
Quote:

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

$\displaystyle \left\{\begin{bmatrix}1\\1\\0\end{bmatrix},\begin{ bmatrix}-1\\0\\1\end{bmatrix}\right\}$
• January 14th 2011, 05:45 PM
Arita
Thanks so much for the help!
• January 14th 2011, 05:59 PM
dwsmith
Quote:

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.