Eigenvalues/Eigenvectors

**Ideasman** · December 11th 2007, 04:43 PM

How do I determine what the original matrix was that yielded these two eigenvalues with the corresponding eigenvectors:

$\lambda_1 = -3$ Eigenvector: [0,1]

$\lambda_2 = 2$ Eigenvector: [1,0]

**Ideasman** · December 11th 2007, 05:23 PM

Originally Posted by Ideasman

How do I determine what the original matrix was that yielded these two eigenvalues with the corresponding eigenvectors:

$\lambda_1 = -3$ Eigenvector: [0,1]

$\lambda_2 = 2$ Eigenvector: [1,0]

Work:

$(\lambda + 3)(\lambda - 2) = \lambda^2 + \lambda - 6$

$(a - \lambda)(d - \lambda) - bc = \lambda^2 + \lambda - 6$

$\lambda^2 - a\lambda - d\lambda + ad - bc = \lambda^2 + \lambda - 6$

I thought the following matrix would work, but it didn't

:

$\left[ \begin {array}{cc} -2&2\\\noalign{\medskip}2&1\end {array}<br /> \right]$

I got the right eigenvalues, but not the right eigenvectors, grr.

**galactus** · December 11th 2007, 05:35 PM

See if this helps:

$det\begin{bmatrix}{\lambda}&1\\{\lambda}+6&{\lambd a}+2\end{bmatrix}={\lambda}^{2}+{\lambda}-6$

The correct matrix will work if $Ax={\lambda}x$

Where x is the eigenvector and $\lambda$ is the eigenvalue.

You are given eigenvalues and eigenvectors.

**Ideasman** · December 11th 2007, 06:13 PM

Originally Posted by galactus

See if this helps:

$det\begin{bmatrix}{\lambda}&1\\{\lambda}+6&{\lambd a}+2\end{bmatrix}={\lambda}^{2}+{\lambda}-6$

The correct matrix will work if $Ax={\lambda}x$

Where x is the eigenvector and $\lambda$ is the eigenvalue.

You are given eigenvalues and eigenvectors.

I tried. I can't figure it out.

EDIT: Yay I figured it out. Good old PDP^(-1). The matrix, incase you were wondering, is [[2,0],[0,-3]]

Thread: Eigenvalues/Eigenvectors

LinkBack

Thread Tools

Display