Finding Jordan form of a real 4x4 matrix, what's wrong with my reasoning?

Question

Let $B=\begin{bmatrix} 0 & 0 & 0 & -8 \\ 1&0&0 & 16\\ 0 &1&0& -14\\ 0&0&1&6 \end{bmatrix}$

Consider B a real matrix. Find its Jordan Form. So, the characteristic polynomial for B is $(x^2-2x+2)(x-2)^2$. Suppose $B$ represents $T$ in the standard basis. By the primary decomposition theorem, we have that $V$ is the direct sum of $K=Ker(T^2-2T+2I)$ and $L=Ker[(T-2)^2]$. $L$ has dimension 2(direct computation here), so that $K$ should have dimension 2, too. Defining an operator $N$ over $Null(T-2)^2$ as $N=T-2$, we have that $N$ is nilpotent(it "dies" when is raised to two), and also $[T]_B=[N]_B+2[I]_B$, for every basis of L. In particular, exists a "cyclic base for N" in L, i.e., a basis for which $N$ can be represented as \begin{bmatrix} 0 & 0 \\ 1 & 0 \\ \end{bmatrix} And then, the matrix of T with respect to that same basis is

\begin{bmatrix} 2 & 0 \\ 1 & 2 \\ \end{bmatrix}

Since the rest can't be factored in linear factors, I guess we can just try to find a basis for K and the matrix will look sthg like this:

$D=\begin{bmatrix} 2 & 0 & 0 & 0 \\ 1&2&0 & 0\\ 0 &0&*& *\\ 0&0&*&* \end{bmatrix}$

The problem is, my friends, that when I compute $B^2-2B+2I$ through WA, and then row reduce it, it says that it is equivalent to the identity matrix, so I can't extract any basis for K. What's wrong here?

Answer 1

You are getting something wrong. The point is, that the eigenvalues of this expression are $2,1+i$ and $1-i$, of which the eigenvalue $2$ has multiplicity $2$, hence the Jordan canonical form $J$ will look like:$$ \begin{pmatrix} 2 \qquad 1\qquad 0\qquad 0 \\ 0\qquad 2 \qquad 0 \qquad 0 \\ 0\qquad 0 \ \quad 1-i \ \quad 0 \\ \ \ 0\qquad 0 \qquad 0 \quad {1+i} \end{pmatrix} $$ This is the Jordan form of the matrix. Now, all you need to do, is to find the following: one eigenvector of $1+i$, one eigenvector of $1-i$ and two linearly independent eigenvectors of $2$.

For this, you need to solve the equations:

$B\vec{x}=(i+1)\vec{x}$

$B\vec{x}=(1-i)\vec{x}$

$B\vec{x}=2\vec{x}$ (will give you two linearly independent solutions)

Once you normalize all the above vectors, then adjoining them as columns will give you the transformation matrix $S$, such that $B=S^{-1}JS$.

Answer 2

The matrix you start with is a companion matrix, so you know the characteristic polynomial and the minimum polynomial is the same. As you have also noticed the characteristic polynomial does not split over the real number field, and it follows that the Jordan form for this matrix does not exist over the real numbers - the Jordan form as given by астон вілла олоф мэллбэрг is correct. The next best thing is the rational canonical form (also known as classical canonical form). $B$ is similar to $$ \begin{bmatrix} 2 & 0 & 0& 0 \\ 1& 2 & 0 & 0 \\ 0& 0 & 0&-2 \\ 0&0&1&2 \end{bmatrix}.$$

Notice that the lower right block is a companion matrix for the polynomial $x^2-2x+2$. For more on the rational canonical form, you can see, for example, this answer.