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Finding the associated unit eigenvector

Mathematics Asked by Laufen on September 30, 2020

Background

Find the eigenvalues $λ_1<λ_2$ and two associated unit eigenvectors of the symmetric matrix

$$A = begin{bmatrix}-7&12\12&11end{bmatrix}$$

My work so far

$$A = begin{bmatrix}-7-λ&12\12&11-λend{bmatrix}=λ^2-4λ-221=(λ+13)(λ-17)$$

Thus

$$λ_1=-13$$
$$λ_2=17$$

To find the solution set for $λ_1$

$$begin{bmatrix}6&12\12&24end{bmatrix}=begin{bmatrix}1&2\0&0end{bmatrix}$$

$$x_1=-2x_2$$
$$x_2=x_2$$
$$=begin{pmatrix}-2\1end{pmatrix}$$
and the solution set for $λ_2$

$$begin{bmatrix}-24&12\12&-6end{bmatrix}=begin{bmatrix}1&frac{-1}{2}\0&0end{bmatrix}$$

$$x_1=frac{1}{2}x_2$$
$$x_2=x_2$$

$$=begin{pmatrix}frac{1}{2}\1end{pmatrix}$$

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 $λ_2$

$$sqrt{(frac{1}{2})^2+1^2}=sqrt{(frac{1}{4})+1}=pmfrac{2}{sqrt{2}}=begin{pmatrix}frac{1}{sqrt{2}}\frac{2}{sqrt{2}}end{pmatrix}$$

I know that I’m off here, but just took a guess.

One Answer

As you correctly found for $lambda_{1}=-13$ the eigenspace is $(−2x_{2},x_{2})$ with $x_{2}inmathbb{R}$. 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.

Correct answer by Gio on September 30, 2020

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