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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.

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  • $\begingroup$ Consider matrices of the form $\lambda I_n$, with $n>1$. They are diagonalizable and only have one eigenvalue. $\endgroup$ – Git Gud Mar 31 '14 at 16:31
  • $\begingroup$ @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? $\endgroup$ – DonAntonio Mar 31 '14 at 16:33
  • $\begingroup$ I realise now I don't even understand what's being asked. $\endgroup$ – Git Gud Mar 31 '14 at 16:36
  • $\begingroup$ @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 $\endgroup$ – George1811 Mar 31 '14 at 16:37
  • $\begingroup$ 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... $\endgroup$ – George1811 Mar 31 '14 at 16:42
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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 $A=\lambda I$.

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