# Lost on rational and Jordan forms

I'm having a lot of trouble trying to understand rational canonical form, primary rational canonical form, and Jordan form. I've looked at the posts about this, but I haven't been able to understand those concepts.

I've been given the following matrix which is associated to an endomorphism $\phi$ of $V=\mathbb{R}^{4}$

$$M=\begin{bmatrix} 1 & 0 & 0 & -1\\ -1 & 0 & -1 & 2\\ 2 & 2 & 2 & -2\\ 0 & 0 & 0 & 1\\ \end{bmatrix}$$

I have to compute the following:

1. Invariant factors of $\phi$ and a base of $V$ for which the matrix associated to $\phi$ is the rational canonical form of $M$.

2. Determine a basis of $V$ for which the matrix associated to $\phi$ is the primary rational canonical form of $M$.

3. The real Jordan canonical form associated to $M$.

Now it comes my try and my doubts:

1. The characteristic polynomial is $(x-1)^2(x^2-2x+2)$ which has just one eigenvalue $(1)$ with algebraic multiplicity $2$. From here, it is easy to see that the minimal polynomial is $(x-1)^2(x^2-2x+2)$ (by multiplying matrixes). However, I've been told that I can also know the minimal polynomial by computing $dim(Nuc(I-M))$ i.e. $(4-rank(I-M))$ and realising it is $1$ i.e. $(4-3)$, but I do not understand that.

Therefore, the only invariant factor is the minimal polynomial, and then the rational canonical form is:

$$C=\begin{bmatrix} 0 & 0 & 0 & -2\\ 1 & 0 & 0 & 6\\ 0 & 1 & 0 & -7\\ 0 & 0 & 1 & 4\\ \end{bmatrix}$$

since $(x-1)^2(x^2-2x+2)= (x^4-4x^3+7x^2-6x+2)$. Now I'm afraid I don't know how to compute the basis I've been asked for. Also, in case that there were two invariant factors, for example, our minimal polynomial and $(x-3)$, I'd know how to write the rational canonical form, but I'm not sure whether the order of the companion matrixes matters.

1. My trouble is quite the same here. I know that the primary rational canonical form is :

$$R=\begin{bmatrix} 0 & -1 & 0 & 0\\ 1 & 2 & 0 & 0\\ 0 & 0 & 0 & -2\\ 0 & 0 & 1 & 2\\ \end{bmatrix}$$

since the elementary divisors are $(x-1)^2$ and $(x^2-2x+2)$ but I'm not sure of the order. And once again, I don't know how to get the basis.

1. Finally, $dim(Nuc(I-M))= 1$ as I just said, so there is 1 jordan block of size at least 1, and therefore, $dim(Nuc(I-M)^2)$ must be $2$ and we have one jordan block for the eigenvalue $1$ and it has size 2.

The problem comes with the block(s) associated to $(x^2-2x+2)$. The jordan form should be $$J=\begin{bmatrix} 1 & 0 & 0 & 0\\ 1 & 1 & 0 & 0\\ 0 & 0 & 1 & -1\\ 0 & 0 & 1 & 1\\ \end{bmatrix}$$

but I don't know why.

Sorry for the long post and thanks in advance.

• For the third question, the polynomial $x^2-2x+2 = (x-(1+i))(x-(1-i))$ yields the eigenvalue $1-i$ and as explained here, you get the bottom-right block \begin{bmatrix} 1 & -1\\ 1 & 1\\ \end{bmatrix} – Watson Dec 30 '15 at 12:43
• @Watson Why $1-i$ and not $1+i$ or both? – Who knows Dec 30 '15 at 13:02
• These complex roots are conjuagted, you just have to take one of them to form the $2 \times 2$ block. If you take $1+i$, then this just swaps the rows of the block. – Watson Dec 30 '15 at 13:07