I was given the following question:
Is $\left(\begin{matrix}1+\sqrt{6}\\-1\\\end{matrix}\right)$ an eigenvector of $\left(\begin{matrix}4&-5\\-1&2\\\end{matrix}\right)$? If so, find the eigenvalue.
I know the way to solve for eigenvalues is to find the determinant $A-\lambda$ of and set it to zero. That gives me eigenvalues of $3-\sqrt{6}$ and $3+\sqrt{6}$. How do I find the eigenvectors from here? Also is there a better way to do this?
(And is there a way to do this on a TI-nspire CAS).
Best Answer
We have $\begin{pmatrix} 4&-5\\-1&2 \end{pmatrix}$ $\begin{pmatrix} 1+\sqrt{6}\\ -1\end{pmatrix}=\begin{pmatrix} 9+4\sqrt{6}\\ -3-\sqrt{6}\end{pmatrix}$.
Is $\begin{pmatrix} 9+4\sqrt{6}\\ -3-\sqrt{6} \end{pmatrix}$ a scalar multiple of $\begin{pmatrix} 1+\sqrt{6}\\ -1\end{pmatrix}$?
Well, certainly it should be $\lambda=3+\sqrt{6}$ (because of the second coordinate) and $(3+\sqrt{6})(1+\sqrt{6})=9+4\sqrt{6}$, so $v$ is an eigenvector of eigenvalue $3+\sqrt{6}$.