Diagonalize a Matrix Example

Process

1. Find eigenvalues
2. Find eigenvectors
3. Perform change of basis with the new set of eigenvectors. Note that this set is linearly independent.

Example

Diagonalize

$$\left[\begin{array}{ccc}0& 3& 0\\ 1& 0& -1\\ 0& 2& 0\end{array}\right]$$

• Find eigenvalues and eigenvectors

$$X=\det \left[\begin{array}{ccc}-\lambda & 3& 0\\ 1& -\lambda & -1\\ 0& 2& -\lambda \end{array}\right]=-{\lambda }^3+\lambda =\lambda (1-\lambda )(1+\lambda )$$

• We get eigenvalues $0,1,-1$
  • For 0-eigenvector, we find a non-zero solution for $(M-0(I))x=0$
    • Then, we get:

$$\left[\begin{array}{cccc}1& 0& -1& 0\\ 0& 1& 0& 0\\ 0& 0& 0& 0\end{array}\right]$$

	- We can pick a $(1,0,1)$ as an eigenvector
- For $1$-eigenvector, we find a non-zero soln for $(M - 1I)x = 0$
	- Then, we get:

$$\left[\begin{array}{cccc}1& 0& -\frac{3}{2}& 0\\ 0& 1& -\frac{1}{2}& 0\\ 0& 0& 0& 0\end{array}\right]$$

	- We can pick a $(3,1,2)$ as an eigenvector
- For $(-1)$-eigenvector, we find a non-zero soln for $(M + 1I)x = 0$
	- We can pick a $(3,-1,2)$ as an eigenvector

• Hence, we have eigenvector-value pairs $(0)(1,0,1),(1)(3,1,2),(-1)(3,-1,2)$
• Now, we perform change of basis. with new basis $\{(1,0,1),(3,1,2),(3,-1,2)\}$

\left[\begin{array}{cc} 1 & 3 & 3\ 0 & 1 & -1\ 1 & 2 & 2\ \end{array}\right]^{-1} \left[\begin{array}{cc} 0 & 3 & 0\ 1 & 0 & -1\ 0 & 2 & 0\ \end{array}\right] \left[\begin{array}{cc} 1 & 3 & 3\ 0 & 1 & -1\ 1 & 2 & 2\ \end{array}\right]

- $$ [T]_{\alpha}^{\alpha'} = \left[\begin{array}{cc} 0 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & -1\\ \end{array}\right]