Finding the associated unit eigenvector
By Sophia Bowman
Background
Find the eigenvalues $λ_1<λ_2$ and two associated unit eigenvectors of the symmetric matrix
$$A = \begin{bmatrix}-7&12\\12&11\end{bmatrix}$$
My work so far
$$A = \begin{bmatrix}-7-λ&12\\12&11-λ\end{bmatrix}=λ^2-4λ-221=(λ+13)(λ-17)$$
Thus
$$λ_1=-13$$$$λ_2=17$$
To find the solution set for $λ_1$
$$\begin{bmatrix}6&12\\12&24\end{bmatrix}=\begin{bmatrix}1&2\\0&0\end{bmatrix}$$
$$x_1=-2x_2$$$$x_2=x_2$$$$=\begin{pmatrix}-2\\1\end{pmatrix}$$and the solution set for $λ_2$
$$\begin{bmatrix}-24&12\\12&-6\end{bmatrix}=\begin{bmatrix}1&\frac{-1}{2}\\0&0\end{bmatrix}$$
$$x_1=\frac{1}{2}x_2$$$$x_2=x_2$$
$$=\begin{pmatrix}\frac{1}{2}\\1\end{pmatrix}$$
However, I'm unsure how to get the associated unit eigenvectors. Would I plug these into the quadratic formula to find the solutions? For example, for $λ_2$
$$\sqrt{(\frac{1}{2})^2+1^2}=\sqrt{(\frac{1}{4})+1}=\pm\frac{2}{\sqrt{2}}=\begin{pmatrix}\frac{1}{\sqrt{2}}\\\frac{2}{\sqrt{2}}\end{pmatrix}$$
I know that I'm off here, but just took a guess.
$\endgroup$ 21 Answer
$\begingroup$As you correctly found for $\lambda_{1}=-13$ the eigenspace is $(−2x_{2},x_{2})$ with $x_{2}\in\mathbb{R}$. So if you want the unit eigenvector just solve:
$(−2x_{2})^2+x_{2}^2=1^2$, which geometrically is the intersection of the eigenspace with the unit circle.
$\endgroup$
