# Generalized Eigenvectors when algebraic multiplicity greater than 1

Find the Generalized Eigenvectors of
$$\begin{bmatrix} 1 & 0 & 0 & 0 & 0\\ 1 & -1 & 0 & 0 & -1\\ 1 & -1 & 0 & 0 & -1\\ 0 & 0 & 0 & 0 & -1\\ -1 & 1 & 0 & 0 & 1 \end{bmatrix}$$

Since for $\lambda = 0, null(A-\lambda I) = 2.$ I have 2 eigenvectors = $$\begin{bmatrix} 0\\0\\0\\1\\0 \end{bmatrix} and \begin{bmatrix} 0\\0\\1\\0\\0 \end{bmatrix}$$ I know to use $(A - \lambda I)v_2 = v_1$ to get all eigenvectors corresponding to a $\lambda$. But this one has 2 blocks corresponding to $\lambda = 0$ in the Jordan Cannonical form. $$J = \begin{bmatrix} 1 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 0 & 1\\ 0 & 0 & 0 & 0 & 0 \end{bmatrix}$$ So, how do I associate generalised eigenvectors to each Jordan block?

• What is generalised eigenvector? – Mostafa Ayaz Jan 29 '18 at 18:44
• Assuming that you know how to calculate general eigenvectors, how long is the Jordan chain for each of the eigenvectors of zero that you’ve got? – amd Jan 30 '18 at 1:06
• For eigenvalue = 0 it's 1 and 3, For eigenvalue = 1 it's 1 – Renganathan Subramanian Jan 30 '18 at 9:52

The first Jordan block is the one-by-one matrix [1]. An associated eigenvector is x = [–1, –1, –1, –1, 1]$^{\mathrm T}.$ The multiplicity of that eigenvalue is one, so there are no complications here.

Dealing with eigenvalue 0 is not as easy. The remainder of my answer is based on Wikipedia subtopic "Computation of generalized eigenvectors."

Because rank(A – 0 I) = rank A = 3, rank A$^2$ = 2, and rank A$^3$ = 1, there will be three linearly-independent generalized eigenvectors—z$_3$ of order 3, z$_2$ of order 2, and z$_1$ of order 1—that form a chain. To find z$_3,$ we solve A$^3$z$_3$ = 0 such that A$^2$z$_3$ ≠ 0. The simplest solution is z$_3$ = [0, 0, 0, 0, 1]$^{\mathrm T}.$ Then z$_2$ = Az$_3$ = [0, –1, –1, –1, 1]$^{\mathrm T},$ and z$_1$ = Az$_2$ = [0, 0, 0, –1, 0]$^{\mathrm T}.$ Those three generalized eigenvectors are associated with the third, three-by-three Jordan block.

Finally, note that z$_1$ is a multiple of the first eigenvector that you give. Hence, you can associate the other eigenvector y = [0, 0, 1, 0, 0]$^{\mathrm T}$ with the second, one-by-one Jordan block [0].

FYI, the Jordan decomposition A = S J S$^{-1}$ has S = [x, y, z$_1,$ z$_2,$ z$_3]$ where the generalized eigenvectors are the columns of S.