Find the rational canonical form of $A=\left(\begin{smallmatrix} -1 & -2 & 6 \\ -1 & 0 & 3\\ -1 & -1 &4 \end{smallmatrix}\right)$ Attempt
First, the characteristic polynomial of $A$ is $\chi_A=(t-1)^3$ and the minimum polynomial is $\mu_A=(t-1)^2$. Since the degree of the minimum polynomial is $2$, the dimension of the largest cyclic subspace cannot exceed 2. Now if $e_1=(1,0,0)^T$  then the cyclic subspace $C_{e_1}$ generated by $e_1$ is $$\mathrm{span}\{e_1, Ae_1\}.$$
Now the space $V$ can be decomposed into a direct sum of cyclic subspaces. so $$V=C_{e_1} \oplus C_x$$ for some $x$. we note that since $V=\mathbb{R}^2$, $C_x$ must be $1$ dimensional. So the minimum polynomial of $A$ restricted to $C_x$ is 1 dimensional. Since this polynomial divides $\mu_A$, we find that the minimum polynomial is $t-1$.  Thus there are two companion matrices, one corresponding to $(t-1)^2$ and one corresponding to $(t-1)$. Thus $$R=\begin{pmatrix} 0 & -1&0\\ 1 & 2 & 0\\ 0 & 0 & 1\end{pmatrix}.$$
Questions and concerns
I want to make sure my reasoning is correct. Throughout the above argument I am implicitly using the assumption that the minimum polynomial of a cyclic subspace of dimension $n$ has degree $n$. Is this correct? In general, can you tell me properties of the rational canonical form of a matrix similar to the first answer here for Jordan canonical form that may help calculate the rational canonical form for matrices of small size?
 A: Your work is fine. However, it is very ad hoc. If you are interested in a more algorithmic way of computing rational canonical forms, read on.
There are actually two things that have the right to be called the "Rational Canonical Form", the one obtained through invariant factors (what most people consider the rational canonical form) and the one obtained from elementary divisors (more akin to the Jordan form). I'll describe how to compute the elementary divisors and then get the invariant factors from them.

Let $T: V \to V$ be a linear transformation.Suppose that $p(x)$ is an irreducible factor of the minimal polynomial of $T$ of degree $d$. We will construct the "dot" diagram corresponding to $p(x)$ which will tell us the elementary divisors corresponding to $p(x)$.
Compute the following numbers until you get $0$:
\begin{align*}
r_1 &= \frac{1}{d}\left( dim(V) - rank(p(T))\right)\\
r_k &= \frac{1}{d}\left( rank(p(T)^{k-1}) - rank(p(T)^k)\right) \text{ for } k \geq 2
\end{align*}
Now construct a diagram with $r_k$ dots in the $k^{th}$ row. The elementary divisors are $p(x)^{s_i}$ where $s_i$ is the number of dots in the $i^{th}$ column of the dot diagram.

Example: Say we are in a real vector space. Suppose $p(x) = x^2+1$ and we compute the $r$ numbers to be $r_1 = 3, r_2 = 3, r_3 = 1, r_4 = 1, r_5 = 0$.
The dot diagram would be:
\begin{array}{lll}
\cdot & \cdot & \cdot \\
\cdot & \cdot & \cdot \\
\cdot \\
\cdot \\
\end{array}
The elementary divisors corresponding to $x^2+1$ would then be $(x^2+1)^4, (x^2+1)^2, (x^2+1)^2$.
To obtain the rational canonical form via elementary divisors, do this for each irreducible factor of the minimal polynomial, and then populate a matrix with the companion matrices of elementary divisors along the diagonal.

To obtain the invariant factors from the elementary divisors: In an array, line up the elementary divisors from greatest degree to least degree with each row corresponding to an irreducible factor of the minimal polynomial. The invariant factors are the products along the columns of this array. The biggest invariant factor is always the minimal polynomial.

For example: Suppose the elementary divisors corresponding to $x^2+1$ are as above, and the elementary divisors corresponding to $x-2$ are: $(x-2)^3, (x-2)$. And that these are the only irreducible factors of the the minimal polynomial.
Then the invariant factors are $$(x^2+1)^4(x-2)^3, (x^2+1)^2(x-2), (x^2+1)^2.$$
To obtain the rational canonical form via invariant factors, populate a matrix with companion matrices of the invariant factors along the diagonal.

In your problem, the only irreducible factor of the minimal polynomial is $(t-1)$. Let's compute its dot diagram.
$$ r_1 = \frac{1}{1}(3 - rank(A - I)) = (3 - 1) = 2 \\
r_2 = \frac{1}{1}(rank(A-I) - rank((A-I)^2)) = (1 - 0) = 1 \\
r_3 = \frac{1}{1}(rank((A-I)^2) - rank((A-I)^3) = (0 -0) = 0$$
So the dot diagram is 
\begin{array}{ll}
\cdot & \cdot \\
\cdot \\
\end{array}
Therefore the elementary divisors are $(t-1)^2, (t-1)$. In this case, since there is only one irreducible factor of the minimal polynomial, the invariant factors are the same as the elementary divisors. Anyway, we get that the rational canonical form (via elementary divisors and invariant factors) is:
$$\begin{pmatrix}
0 & -1 & 0 \\
1 & 2 & 0 \\
0 & 0 & 1
\end{pmatrix}$$
just as you had.
A: Your reasoning is correct, in particular the minimal polynomial of a cyclic subspace has degree equal to the dimension of that space. This is basically just the fact that for a monic polynomial $P$ the $K[X]$-module $K[X]/(P)$, which is the model for a cyclic subspace with minimal polynomial $P$, has $K$-dimension $\deg P$.
However your argument could be shortened. Given that the minimal polynomial is $(X-1)^2$, you know that this is the final invariant factor. The degrees of the invariant factors must add up to $3$, so you need another factor of degree$~1$, and since it must divide the minimal polynomial it can only be equal to$~X-1$. This gives you your two companion-matrix diagonal blocks, and hence the rational canonical form. The definition of the RCF as I know it puts the blocks in order of the invariant factors, where each divides the next one; that gives
$$
R=\pmatrix{1&0&0\\0&0&-1\\0&1&2}.
$$
