question regarding generalized eigenvectors for Differential equation

Question

I also have a question about the generalized eigenvector, I learned how to find the generalized vector in class, but for this practice problem, I am not sure how to find the generalized eignvectors for a 3 by 3 matrix.

here is the image of the question

My approach is the following:

Since the matrix A is upper triangular, which means the eigenvalues would just be the diagonal entries, and here we have repeated root -1 with the algebraic multiplicity of 2, and another root of 2. And I plug the eignvalue -1 and performing the row-reduced echelon form, I found the final matrix to be

\begin{bmatrix}0&1&0\\0&0&1\\0&0&0\end{bmatrix}

and I am not sure how to go from here to find the generalized eigenvecotors, I guess my question would be how do you find the kernel and using kernel to find the generalized eigenvectors.

Any help would be greatly appreciated.

Thank you so much!

Answer 1

You don't need to complicate yourself with kernels and stuff, first of all a kernel is an entity used in linear transformations. The procedure is rather straight-forward and simple :

Since $λ_1= \pm 1$ is an eigenvalue of multiplicity $2$ of the matrix $A$, what you need to do to find a generalized eigenvector is :

• Find a "common" eigenvector $v_1$, by solving the system : $(A-λ_1I)v_1=0$
• Solve the system : $(A-λ_1I)v_2=v_1$, to find the generalised eigenvector $v_2$.

The eigenvalue $λ_2=2$ is of multiplicity $1$ and does not require a generalised eigenvector.

Hope that cleared your mind a little bit, if you got any questions ask down below !