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

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

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

Printable View

- Dec 11th 2007, 04:43 PMIdeasmanEigenvalues/Eigenvectors
How do I determine what the original matrix was that yielded these two eigenvalues with the corresponding eigenvectors:

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

$\displaystyle \lambda_2 = 2$ Eigenvector: [1,0] - Dec 11th 2007, 05:23 PMIdeasman
Work:

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

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

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

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

$\displaystyle \left[ \begin {array}{cc} -2&2\\\noalign{\medskip}2&1\end {array}

\right]$

I got the right eigenvalues, but not the right eigenvectors, grr. - Dec 11th 2007, 05:35 PMgalactus
See if this helps:

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

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

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

You are given eigenvalues and eigenvectors. - Dec 11th 2007, 06:13 PMIdeasman