0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Seeing the block structure you can maybe derive what the corresponding "block basis" would be. $\endgroup$ – mathreadler Sep 14 '17 at 21:40
  • $\begingroup$ @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, $v_1 = [ i,1,i,1]^T$. $\endgroup$ – TunaBooties Sep 15 '17 at 4:19
1
$\begingroup$

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 $(0,0,i,1)$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.