$$

Find an orthonormal basis composed of eigenvectors

Problem

Let

$$A=\left[\begin{array}{cccc}0& 0& 1& 1\\ 0& 0& 1& 1\\ 1& 1& 0& 0\\ 1& 1& 0& 0\end{array}\right].$$

Find the eigenvalues of

$A$ and an orthonormal basis of

${\mathbb{R}}^4$ composed of eigenvectors of

$A$.

Thought

Since

$A$ is a symmetric matrix, its eigenvalues are real and there is an orthonormal basis composed of eigenvectors of

$A$ by the spectral theorem. Thus, the process of finding eigenvalues are the same, but we have to solve for an orthonormal basis of

$\ker (A-\lambda I)$ for each eigenvalue

$\lambda$. Eventually, we may normalize the length of all eigenvectors to make it orthonormal.

Sample answer

One may compute the characteristic polynomial

$$p_A(x)=x^4-4x^2.$$

Thus, the eigenvalues are

$2,-2,0,0$.

For

$\lambda =2,-2$, one may find eigenvectors

$${\mathbf{u}}_1=\left[\begin{array}{c}1\\ 1\\ 1\\ 1\end{array}\right],\text{ and }{\mathbf{u}}_2=\left[\begin{array}{c}1\\ 1\\ -1\\ -1\end{array}\right].$$

For

$\lambda =0$, we may follow Find an orthonormal basis of a subspace to find an orthogonal basis of

$\ker (A-0I)$ and get

$${\mathbf{u}}_3=\left[\begin{array}{c}1\\ -1\\ 0\\ 0\end{array}\right],\text{ and }{\mathbf{u}}_4=\left[\begin{array}{c}0\\ 0\\ 1\\ -1\end{array}\right].$$

By replacing

${\mathbf{u}}_i$ with

$\frac{1}{\Vert {\mathbf{u}}_i\Vert }{\mathbf{u}}_i$, we have

$${\mathbf{u}}_1=\frac{1}{2}\left[\begin{array}{c}1\\ 1\\ 1\\ 1\end{array}\right],{\mathbf{u}}_2=\frac{1}{2}\left[\begin{array}{c}1\\ 1\\ -1\\ -1\end{array}\right],{\mathbf{u}}_3=\frac{1}{\sqrt{2}}\left[\begin{array}{c}1\\ -1\\ 0\\ 0\end{array}\right],\text{ and }{\mathbf{u}}_4=\frac{1}{\sqrt{2}}\left[\begin{array}{c}0\\ 0\\ 1\\ -1\end{array}\right].$$

Finally,

$\{{\mathbf{u}}_1,{\mathbf{u}}_2,{\mathbf{u}}_3,{\mathbf{u}}_4\}$ is an orthonormal basis of

${\mathbb{R}}^4$ composed of eigenvectors of

$A$.

This note can be found at Course website > Learning resources.