Jordan Form

Jordan form captures the case when the Transformation Matrix $Q$ is not invertible (eigenvectors are not linearly independent), and we can achieve an almost Diagonal Form.

---

Transformation Matrix - $T=Q^{-1}$ → $\bar{x}=Q^{-1}x$

The core idea is the following two equations:

$$\boxed{\begin{array}{rl}A{v}_1& =v_1{\lambda }_1\\ A{v}_2& =v_1\cdot 1+v_2{\lambda }_1\end{array}}$$

---

Given a repeated eigenvalue, ${\lambda }_r$, we have two different options depending on the Nullity of ${\lambda }_r\mathbb{I}-A$.

Nullity larger than 1

• We can find several linearly independent solutions Nullity less than the repetitions of the eigenvalue
• Aka we dont have sufficient distinct eigenvectors. In this case we use a Jordan Block

Example

Given the system matrix

$$A=\left[\begin{array}{cc}0& -1\\ 0& -2\end{array}\right]$$

The eigenvalues are

$$\begin{array}{rl}\det (\lambda \mathbb{I}-A)& =0\\ {\lambda }_{1,2}& =-1\end{array}$$

And the eigenvector is

$$\begin{array}{rl}(\lambda \mathbb{I}-A)q& =0\\ q& =\left[\begin{array}{c}1\\ 1\end{array}\right]\end{array}$$

The core idea is now the following two equations

$$\boxed{\begin{array}{rl}A{v}_1& =v_1{\lambda }_1\\ A{v}_2& =v_1\cdot 1+v_2{\lambda }_1\end{array}}$$

Giving us $Q=[v_1{v}_2]=\left[\begin{array}{cc}1& \frac{1}{2}\\ 1& -\frac{1}{2}\end{array}\right]$.