# How to find generalized Eigen vectors of a matrix with Eigen vectors already on diagonal?

I have a matrix

A= \begin{bmatrix} 1 & -1 & 0 & 0 \\ 1 & 1 & 0 & 0 \\ 0 & 0 & 1 & -1\\ 0 & 0 & 1 & 1 \end{bmatrix}

This matrix has eigen values $\lambda = 1+i, 1-i$ where each have multiplicity 2.

Notice this matrix is also already is the form: $A= diag(\begin{bmatrix} a_j & -b_j \\ b_j & a_j \end{bmatrix})$

Where $\lambda = a_j + ib_j$ for $\lambda_1 = 1+i$. In other words, this matrix already is diagonalized with the eigen values on the diagonal.

For the problem, I am trying to find the eigen vectors and corresponding generalized eigen vectors for the eigen values to make a basis of $\mathbb{R}^4$, by using the real and imaginary parts of the eigen vector and generalized eigen vector for $\lambda_1 = 1+i$.

I have found the eigen vector for $\lambda_1 = 1+i$ to be $v_1 = \begin{bmatrix} i \\ 1 \\ i\\ 1 \end{bmatrix}$.

But when I then try to find the generalized eigen vector for this eigen value by solving the system:

$(A-\lambda_1I)^2v_2 = {0}$

The system row reduces to the same system that gave me $v_1$, and I again get $v_1$ as my generalized eigen vector, which does not provided me with a new linearly independent vector to use for my basis. I think this has to do with the matrix A already having the eigen values on the diagonal but I am not sure.

Ref. pg 36 and 38 #3e of Lawrence Perko's, Differential Equations and Dynamical Systems.

• Seeing the block structure you can maybe derive what the corresponding "block basis" would be. – mathreadler Sep 14 '17 at 21:40
• @Moo But why is it that you can split the original eigenvector that I found into these two new vectors? I thought solving the system would give the eigenvector that I found, $v_1 = [ i,1,i,1]^T$. – TunaBooties Sep 15 '17 at 4:19

Since the matrix is block diagonal, an eigenvalue and the corresponding eigenvector provided by one block becomes the eigenvalue of the whole matrix, and the block eigenvector, stacked with $0$ to complete the dimension, will become the eigenvector of the whole matrix.
So in your case, to the eigenvalue $1+i$, for instance corresponds the eigenvectors $(i,1,0,0)$ and $(0,0,i,1)$.