# Computation of product of generalized eigenspace and eigenvectors

My question concerns why $(A-\lambda_1 I)(C_1 v_1 + C_2 v_1^{(1)}) = C_2 v_1$ in the calculations below.

Suppose we have a linear system of differential equations with a constant matrix $A \in \mathbb{R}^{3x3}$. $A$ has two distinct eigenvalues $\lambda_1,\lambda_2$ of which the $\lambda_1$ has algebraic multiplicity $n_j =2$ while $\lambda_2$ has $n_j =1$. The general solution can then be found as

$x(t) = e^{At}x_0 = \sum_{j=1}^{2}\left(\left[ \sum_{k=0}^{n_j-1} (A-\lambda_j I)^k \frac{t^k}{k!} \right] x^{0,j} e^{\lambda_j t}\right)$

Choosing $x_0 = \sum_{j=1}^{2} x^{0,j} = C_1v_1 + C_2v_1^{(1)}+C_3v_2$ we get

$x(t) = e^{\lambda_1 t}(C_1v_1 + C_2v_1^{(1)}) + te^{\lambda_1t}(A-\lambda_1 I)(C_1v_1 + C_2v_1^{(1)}) + C_3 e^{\lambda_2 t} v_2 =\\= e^{\lambda_1 t}(C_1v_1 + C_2v_1^{(1)}) + te^{\lambda_1t}(C_2v_1) + C_3 e^{\lambda_2 t} v_2$

What I don't understand is why $(A-\lambda_1 I)(C_1 v_1 + C_2 v_1^{(1)}) = C_2 v_1$.

## 1 Answer

This is by the definition of generalised eigenvectors. In your case, $v_1$ is a generalised eigenvector of rank 1 (which is basically the same as an ordinary eigenvector). This means that it obeys the equation \begin{equation} A v_1 = \lambda_1 v_1 \qquad \Rightarrow \qquad (A - \lambda_1 I) v_1 = 0. \end{equation} Generalised eigenvectors form a Jordan chain. In this case, this means that the generalised eigenvector $v_1^{(1)}$, which is of rank 2, obeys the equation \begin{equation} (A - \lambda_1 I)v_1^{(1)} = v_1, \end{equation} such that \begin{equation} (A-\lambda_1 I)^2 v_1^{(1)} = 0. \end{equation}

• Thank you @frits-veerman, that makes a lot of sense! – langkilde Mar 31 '16 at 12:02