In my dynamical systems, we are asked to find the Jordan Canonical form of the Jacobian in order to analysis the linear stability at fixed points in a second order system. I believe that even for one real degenerate eigenvalue you can admit two linearly independent eigenvectors and therefore diagonalize the matrix into a diagonal matrix. However the lecture notes say to have the $1$ in the upper right hand corner. Who is right? Here's what the lectures notes say:
One real degenerate eigenvalue: $\lambda_1=\lambda_2=\lambda \in \mathbb{R}$. In this case the corresponding Jordan form is $$J^*=\begin{pmatrix} \lambda & 1 \\ 0 & \lambda \end{pmatrix}$$ i.e. we have the single eigenvalue on the digonal, and a $1$ in the upper right hand corner. Note: An exception occurs when $J=\lambda I$ i.e. when $J$ is proportional to the unit matrix as in this case $J^*=P^{-1}JP=J$ whatever $P$.
Now she has included the exception however a matrix with a repeated eigenvalue can still be diagonalizable can it not? I was led to believe that only when the matrix cannot be reduced to a diagonal form then to use the above Jordan form.
Thanks.
Best Answer
Consider a $2 \times 2$ matrix $A$ with a single eigenvalue $\lambda$. For the eigenspace associated with $\lambda$ to be dimension two, $(\lambda I-A)$ must be the zero matrix (you need two independent parameters). So this is only possible if $A=\lambda I$.