Jordan canonical form over $\mathbb{R}$

Question

The theorem about Jordan canonical form states that for any operator $f:V\to V$ where $V$ is a vector space over $\mathbb{C}$ there is a basis such that the matrix of this operator in this basis is a union of Jordan blocks.

And in my lecture notes there is a remark that the theorem is not true for vector spaces over $\mathbb{R}$. After some time I came up with an example: indeed let's take an operator $f:\mathbb{R}^2\to \mathbb{R}^2$ given by matrix $$A_f=\begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}.$$

Suppose it is true and $J$ is a Jordan canonical form then there is a matrix $C$ such that $\det C\neq 0$ with $J=C^{-1}A_fC$ or $CJ=A_fC$.

But $J$ can be one of the following options: $\begin{bmatrix} \lambda & 0 \\ 0 & \lambda \end{bmatrix},\begin{bmatrix} \lambda & 1 \\ 0 & \lambda \end{bmatrix}, \begin{bmatrix} \lambda & 0 \\ 0 & \mu \end{bmatrix} (\lambda\neq \mu)$. In other words, we can have two block $J_1(\lambda)$, one block $J_2(\lambda)$ to two blocks with different diagonal elements.

If $C=\begin{bmatrix} a & b \\ c & d \end{bmatrix}$ then comparing elements of the first column in matrix equation $CJ=A_fC$ we have: $$a\lambda=-c \quad \text{and} \quad c\lambda=a.$$ Since $\det C\neq 0$, i.e. $ad\neq bc$ then it is easy to claim that $a\neq 0$. Then $\lambda=-\frac{c}{a}$ and using the second equality we have $\dfrac{-c^2}{a}=a$ or $a^2+c^2=0$. And since $a,c$ are reals then $a=c=0$ which is contradiction.

Is this reasoning correct? Would be very grateful for any remarks.

Answer 1

Your reasoning is correct - you can somewhat simplify your reasoning by first noting that you can only use two cases - either there is a block of size $2$, or there are two blocks of size one (possibly with the same eigenvalue). Then, starting from the equations $$a\lambda=-c$$ $$c\lambda = a$$ you can work things out more simply by not dividing. Multiplying the first equation by $\lambda$ gives $$a\lambda^2=-c\lambda=-a$$ where we use the second equation to substitute for the second equality. This either implies that $a=0$ or that $\lambda^2=-1$ - the former being a contradiction, as you note, because then $C$ is not invertible, and the latter not being true of any real $\lambda$.

A much easier and more conceptual argument would be just to observe that your matrix $A_f$ does not have any (non-zero) eigenvectors, but $J$ certainly does - and hence they cannot be conjugate. This is a bit closer to the sort of reasoning that is involved in constructing something like the Jordan canonical form - and it makes it more obvious why the equation $\lambda^2=-1$ should show up: it's because that's, essentially, coming from the characteristic polynomial of $A_f$, which has no real roots. Generally, it's a good idea to avoid matrices when doing theoretical work, since, while it works out for a $2\times 2$ case like this, such work is harder to extend than work based upon basis-free reasoning.