# Jordan Normal Forms System of Differential Equations

Investigate the two-dimensional linear system $\begin{bmatrix} x' \\ y' \end{bmatrix} = A \begin{bmatrix} x \\ y \end{bmatrix}$ where $A = \begin{bmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{bmatrix}$ for the case when $D=det(A)=0$ and determine the normal forms that arise.

The characteristic polynomial is $P(\lambda) = det(A-\lambda I) = \lambda^2 - S\lambda +D$ with $S= trace(A)= a_{11}+a_{22}$. Then for $D=det(A)=0$, the characteristic polynomial becomes $P(\lambda) = det(A-\lambda I) = \lambda^2 - S\lambda = \lambda(\lambda - S)$ so we get eigenvalues $\lambda = 0$ and $\lambda = S$.

I am not sure how to proceed to find the Normal Forms.

There are two possibilities to consider: $S=0$ and $S\ne0$.
Taking the second possibility first, you’ve got a $2\times2$ matrix with 2 distinct eigenvalues. The possibilities for the Jordan normal form of such a matrix are pretty limited. It’s going to have two Jordan blocks, obviously, so what does this mean for the sizes of those blocks?
If $S=0$ there again two possibilities to consider: either $A$ is diagonalizable, in which case $A=0$ (can you see why?) or it is not. In the latter case, how many Jordan blocks does this matrix need and what are their sizes? The answers will whittle down the possibilities for the normal form quite dramatically.
• @SamWhelchel no, since that would make the matrix at least $4\times4$. – amd Apr 4 '17 at 5:42
• I think it would be the form $\begin{bmatrix} \lambda & 0 \\ 0 & \mu \end{bmatrix}$ where $\lambda$ and $\mu$ are the eigenvalues, but I am still messing with the matrix. – Sam Whelchel Apr 5 '17 at 1:35