# Help decoding my professor's comments on Jordan canonical form and its relation to systems of linear ODEs

Note: I was planning on posting this as a question, but in typing this up I feel as though I understand most of what's going on here (funny how that happens). That said, I'd love to get some input on whether I am correct, or if there are things I've missed

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My ODE professor has not been very helpful with regard to this topic, so I am hoping that math.stackexchange can help out here.

My professor has been speaking about "Jordan canonical form" and its relation to a system of linear ODEs. We have been studying a general system given by $$\frac{d\vec{x}}{dt} = A\vec{x}$$ (where $A$ is a $2$x$2$ matrix in our case). If we can diagonalize $A$ (I'm not sure why we can assume that this is possible, but I'll assume we can), then by a similarity transformation, we get $$J = T^{-1}AT$$ where $J$ is a diagonal matrix with eigenvalues of $A$ on the diagonal, and $T$ is a square matrix with eigenvectors of $A$ for columns.

Here is where I start to lose understanding of what's going on: Next he says that we can "use a change of coordinates" and let $$\vec{x}(t) = T\vec{z}(t) \iff \vec{z}(t)=T^{-1}\vec{x}(t)$$ so we can rewrite our initial system as $$T\frac{d\vec{z}}{dt} = AT\vec{z} \iff \frac{d\vec{z}}{dt} = T^{-1}AT\vec{z} \iff \frac{d\vec{z}}{dt} = J\vec{z}$$ Since $J$ is diagonal with constants, we can easily solve for $\vec{z}(t)$, and then multiply by $T$ to get back the actual solution, $\vec{x}(t)$.

Case 2: If $A$ is not diagonalizable, then we can instead let $T$ be a matrix with an eigenvector for one column, and a generalized eigenvector, $\vec{y}$, as the other column, where $\vec{y}$ solves $$(A-\lambda I)^2\vec{y} = \vec{0}$$ If this is the case, then we will have (through some matrix operations that I haven't worked out yet) $$J=\begin{bmatrix} \lambda & 1 \\ 0 & \lambda \end{bmatrix}$$ in which case the system $$\frac{d\vec{z}}{dt} = J\vec{z}$$ is still not too hard to solve.

Case 3: If eigenvalues are complex, then if we have $$\lambda_{1,2} = \alpha \pm \beta i$$, then $$J= \begin{bmatrix} \alpha & \beta \\ -\beta & \alpha \end{bmatrix}$$

If this is the case, then I'm not sure how we would solve $$\frac{d\vec{z}}{dt} = J\vec{z}$$ In class he did an example in which he switched to polar coordinates, but it seemed more like a special case...

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If you've made it this far, thanks a bunch, as I said above, I'd love any comments on whether this is correct, or if there is anything I should add.

• @WillJagy Oops. Wrong tag. – Alex Oct 29 '16 at 22:12
• It isn't at all clear what your question is here. At a guess, I'd say that you're unfamiliar with Jordan Canonical form and also perhaps uncomfortable working with complex exponentials (with which you can solve case 3.) If you have a more specific question, then it would be wise to ask it. – Brian Borchers Oct 29 '16 at 22:37