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This is what my professor wrote:

Let \begin{equation*} y_i = y_i(x) ~~~(i = 1, 2, 3) \end{equation*} be differentiable functions in $x$. Set \begin{equation*} Y = \begin{pmatrix} y_1\\ y_2\\ y_3 \end{pmatrix} , Y^{'} =\begin{pmatrix} y_{1}^{'}\\ y_{2}^{'}\\ y_{3}^{'} \end{pmatrix} \end{equation*} We will solve the following differential equations, where $A = (a_{ij})$ is in $M_3(\mathbb{R})$: \begin{equation*} \begin{pmatrix} y_{1}^{'}\\ y_{2}^{'}\\ y_{3}^{'} \end{pmatrix} = Y^{'} = AY = \begin{pmatrix} a_{11} &a_{12} &a_{13}\\ a_{21} &a_{22} &a_{23}\\ a_{31} &a_{32}& a_{33} \end{pmatrix} \begin{pmatrix} y_1\\ y_2\\ y_3 \end{pmatrix} , \end{equation*} or equivalently \begin{eqnarray*} y_{1}^{'}&=& a_{11}y_1 + a_{12}y_2 + a_{13}y_3\\ y_{2}^{'} &= &a_{21}y_1 + a_{22}y_2 + a_{23}y_3\\ y_{3}^{'} &= &a_{31}y_1 + a_{32}y_2 + a_{33}y_3. \end{eqnarray*} Let $P$ be an invertible matrix in $M_3(\mathbb{R})$ such that \begin{equation*} J := P^{-1} A P \end{equation*} is a Jordan canonical form of $A$. etc etc etc...

My question (could be trivial) is that i thought that the matrix $A$ taken over $\mathbb{R}$ is not guaranteed to have a Jordan canonical form. Or am i missing something?

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The Jordan canonical form of a $3\times 3$ matrix can be a diagonal or not.

If not, there are two cases:

  1. There's one block of size 1 with an eigenvalue and a block of size 2 with an eigenvalue in the diagonal and a $1$ in the right upper corner of the block.
  2. There's one block of size $1$ with an eigenvalue and a block of size 2 of the form $$\begin{pmatrix}a &b\\ -b & a \end{pmatrix}$$ The coefficients $a,b$ are the real and imaginary parts of the complex eigenvalue. You can read this here.

It's guaranteed that we have a block of size $1$, because the characteristic polynomial has degree $3$ and a real root.

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