# How to determine generalized eigenvectors of $\begin {bmatrix} 2 & 1 & 0 & 0 \\ 0 & 2 & 0 & 0 \\ 0 & 0 & 2 &1 \\ 0 & 0 & 0 & 2 \end{bmatrix}$

I want to calculate the general solution of this DE-system:

$$\frac{d \vec x}{d t}= A \vec x,\text{ with }A = \begin {bmatrix} 2 & 1 & 0 & 0 \\ 0 & 2 & 0 & 0 \\ 0 & 0 & 2 &1 \\ 0 & 0 & 0 & 2 \end{bmatrix}$$

$\lambda=2$ is eigenvalue with algebraic multiplicity $4$.

Calculating Eigenvectors:

$$\begin {bmatrix} 2-2 & 1 & 0 & 0 \\ 0 & 2-2 & 0 & 0 \\ 0 & 0 & 2-2 &1 \\ 0 & 0 & 0 & 2-2 \end{bmatrix} = \begin {bmatrix} 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 &1 \\ 0 & 0 & 0 & 0 \end{bmatrix}$$

So eigenvectors are obviously
$$\vec v_1 = \begin {bmatrix} 1 \\ 0 \\ 0 \\ 0 \end{bmatrix}\text{ and }\vec v_2= \begin {bmatrix} 0 \\ 0 \\ 1 \\ 0 \end{bmatrix}.$$

My question is, how to calculate generalized eigenvectors $\vec v_3$ and $\vec v_4$ in this case.

Would it be correct, to solve the following to two linear system?

(1) $(A-\lambda I)\vec v_3 = \vec v_1$

(2) $(A-\lambda I)\vec v_4 = \vec v_2$

• $({\bf A}-\lambda {\bf I}){\bf v_3}$ being a linear combination of only $\bf v_3$ and $\bf v_1$, then $\bf v_3$ is in the same generalized eigenspace as $\bf v_1$ – mathreadler May 23 '15 at 21:58
• Sorry if not conforming to your notation, I'm used to writing matrices large letter bold and vectors small letter bold. – mathreadler May 23 '15 at 21:59
• this is already in jordan canonical form. you can compute the $e^{At}$ directly without computing eigenvectors and generalized eigenvectors. – abel May 23 '15 at 22:43

The system you propose is fine. You should find the following generalized eigenvectors: $$\vec{v}_3=\begin{bmatrix}0\\1\\0\\0\end{bmatrix},\quad\vec{v}_4=\begin{bmatrix}0\\0\\0\\1\end{bmatrix}$$
• So the general solution is $x(t)=c_1e^{2t}\vec v_1+c_2e^{2t}\vec v_2+c_3e^{2t}(\vec v_3+t\vec v_1)+c_4e^{2t}(\vec v_4+t\vec v_2), c_1,c_2,c_3,c_4 \in \mathbb{R}$ – fear.xD May 23 '15 at 21:37
So you would want to try and solve $({\bf A}-\lambda {\bf I})^k{\bf v} = {\bf 0}$, but $({\bf A}-\lambda {\bf I})^{k-1}{\bf v} \neq {\bf 0}$ to answer your question. You can use the first order eigenvectors (ordinary eigenvectors) together with the block-zero property to limit the span of generalized eigenvectors. Then you will know you only need to search the subspaces $(a_1,a_2,0,0)^t$ and $(0,0,b_1,b_2)^t$.
• So solving $(A-\lambda I)^2\vec v_3=0$. But $(A-\lambda I)^2$ is the zero matrix. Can i choose $\vec v_3$ (and $\vec v_4$) as i like? – fear.xD May 23 '15 at 21:18