# Finding the eigenvector

• Jun 10th 2009, 01:56 PM
Nerdfighter
Finding the eigenvector
Given a specific Eigenvalue $\lambda$, how do I find an eigenvector of a Matrix $A$?

Specifically, if $\lambda = 4$ and $A =$ $\left[\begin{array}{rrr}3&0&-1\\2&3&1\\-3&4&5\end{array}\right]$

By the way, is there a way to make my Matrix look a little neater/lined up?
• Jun 10th 2009, 02:07 PM
TheAbstractionist
If $\begin{pmatrix}x\\y\\z\end{pmatrix}$ is an eigenvector, then $\left(\begin{array}{rrr}3&0&-1\\2&3&1\\-3&4&5\end{array}\right)\begin{pmatrix}x\\y\\z\end{ pmatrix}\ =\ \lambda\begin{pmatrix}x\\y\\z\end{pmatrix}\ =\ 4\begin{pmatrix}x\\y\\z\end{pmatrix}.$

There will be infinitely many eigenvectors corresponding the the eigenvalue 4, so just pick one of them.

Quote:

Originally Posted by Nerdfighter
By the way, is there a way to make my Matrix look a little neater/lined up?

Instead of centre-aligning them, right-align them (as I have done).
• Jun 10th 2009, 02:17 PM
Nerdfighter
How Do I pick one? I obviously can't pick any numbers for x, y, and z.
• Jun 10th 2009, 02:44 PM
TheAbstractionist
I did not say pick any numbers for $x,y,z.$ Pick $x,y,z$ so that they satisfy the eigenvalue equation.
• Jun 10th 2009, 02:46 PM
Nerdfighter
Quote:

Originally Posted by TheAbstractionist
I did not say pick any numbers for $x,y,z.$ Pick $x,y,z$ so that they satisfy the eigenvalue equation.

What is a good method to find $x,y,z$ such that $x,y,z$ satisfies the eigenvalue equation?

Could I simply pick the zero vector to be an eigenvector?
• Jun 10th 2009, 03:02 PM
TheEmptySet
Quote:

Originally Posted by Nerdfighter
Given a specific Eigenvalue $\lambda$, how do I find an eigenvector of a Matrix $A$?

Specifically, if $\lambda = 4$ and $A =$ $\left[\begin{array}{rrr}3&0&-1\\2&3&1\\-3&4&5\end{array}\right]$

By the way, is there a way to make my Matrix look a little neater/lined up?

Solve the matrix equation

$A\vec v = \lambda \vec v \iff (A-\lambda I)\vec v =0$

Since your eigen value is 4 we get

$(A-4 I)\vec v =0$

$\begin{bmatrix}
3-4 && 0&& -1\\
2 && 3-4 && 1 \\
-3 && 4 && 5-4 \\

\end{bmatrix}$

$\begin{bmatrix}
-1 && 0&& -1\\
2 && -1 && 1 \\
-3 && 4 && 1 \\

\end{bmatrix}$

putting this in reduced row form we get

$\begin{bmatrix}
1 && 0&& 1\\
0 && 1 && 1 \\
0 && 0 && 0 \\

\end{bmatrix}$

Solving the above homogenious equation gives

$z=t, y=-t,x=-t$ or

$\begin{bmatrix}
-t \\ -t \\ t
\end{bmatrix}=t \begin{bmatrix}
-1 \\ -1 \\ 1
\end{bmatrix}$

So that is the eigen vector

p.s if you use the \begin{bmatrix} \end{bmatrix} tags it will auto format all of your columns
• Jun 10th 2009, 03:06 PM
Nerdfighter
Thank you, that's what I needed!
• Jun 10th 2009, 10:41 PM
Swlabr
Quote:

Originally Posted by Nerdfighter
Could I simply pick the zero vector to be an eigenvector?

No, the zero vector is never an Eigenvector, as if it was every scalar would be an Eigenvalue. An Eigenvector for the corresponding Eigenvalue $\lambda$ is defined to be a "non-zero vector such that $Mv = v \lambda$".
• Jun 10th 2009, 10:42 PM
Nerdfighter
Quote:

Originally Posted by Swlabr
No, the zero vector is never an Eigenvector, as if it was every scalar would be an Eigenvalue. An Eigenvector for the corresponding Eigenvalue $\lambda$ is defined to be a "non-zero vector such that $Mv = v \lambda$".

Ooh, very good to know. Thank you.