# Jordan canonical forms and diagonalizing.

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.

• Consider matrices of the form $\lambda I_n$, with $n>1$. They are diagonalizable and only have one eigenvalue. – Git Gud Mar 31 '14 at 16:31
• @George, you're right: also with a single eigenvalue a matrix can be diagonalized, as Git's comment shows. But perhaps the term "degenerate eigenvalue" means the matrix can not bediagonalized? – DonAntonio Mar 31 '14 at 16:33
• I realise now I don't even understand what's being asked. – Git Gud Mar 31 '14 at 16:36
• @DonAntonio en.wikipedia.org/wiki/Degeneracy_%28mathematics%29 - a degenerate eigenvalue (i.e. a multiply coinciding root of the characteristic polynomial) is one that has more than one linearly independent eigenvector. Taken from wikipedia, seems to imply matrices with degenerate eigenvalues $\textbf{can}$ be diagonalized – George1811 Mar 31 '14 at 16:37
• However from wolfram alpha -"If the eigenvalues are n-fold degenerate, then the system is said to be degenerate and the eigenvectors are not linearly independent." -mathworld.wolfram.com/Eigenvalue.html Lol I have no idea... – George1811 Mar 31 '14 at 16:42

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 $A=\lambda I$.