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]
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.
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.