Write the Jordan form of an operator

Question

These are the properties that apply to the operator $A$.

$k_A(x)=x^4(x-2)^4, d(A)=2, d(A^2)=4, d((A-2I))=2, (d((A-2I)^2)=3$

$d$ denotes the defect. $k_A$ is the characteristic polynomial.

I honestly don't know where to go from here.

Answer 1

First, $\;A\;$ is a $\;6\times 6\;$ matrix as its char. polynomial is of degree six. Second, it has only two eigenvalues: zero, of multiplicity four, and $\;2\;$ of multiplicity two. This already means that in the matrix's Jordan Canonical form there'll be four zeros and two twos on the main diagonal.

If I understoof correctly your notation, $\;d(A)=2\iff \text{rank}\,A=4\;$ , and since any Jordan block of the zero eigenvalue diminishes by one the rank of the matrix, this means there are exactly two such block of size two each, so that the block for zero is:

$$\begin{pmatrix}0&1&0&0\\0&0&0&0\\0&0&0&1\\0&0&0&0\end{pmatrix}$$

We're thus left with the block for $\;2\;$, which can be two $\;1\times 1\;$ or one of size $\;2\times2\;$

But note the above block is nilpotent of degree two, so it'll vanish when we square our matrix, and we're given $\;d(A-2I)=2\iff \text{rank}\,(A-2I)=4\;$ , so there can not be one single block for the eigenvalue $\;2\;$ (why?) and it thus must split in two $\;1\times 1\;$ blocks, leading us to

$$\begin{pmatrix}\color{red}0&\color{red}1&0&0&0&0\\\color{red}0&\color{red}0&0&0&0&0\\0&0&\color{red}0&\color{red}1&0&0\\0&0&\color{red}0&\color{red}0&0&0\\0&0&0&0&2&0\\0&0&0&0&0&2\end{pmatrix}$$

Answer 2

let me see if i can work this out. if you have trouble following refer to my early answer for concreteness.

remember the $0 \subset N(A) \subset N(A^2) = N(A^3) \ldots.$ draw a smaller box for $N(A)$ and a bigger box containing $N(A)$ for the $(N(A^2).$ find two linearly independent vectors $u_1, u_2$ in bigger box(possible because $dim(N(A^2) - \dim(N(A)) = 2.$ the vectors $Au_1$ and $Au_2$ is now in the smaller box. that is $u_1, Au_1$ and $u_2, Au_2$ form two chains of length 2 a basis fomr the generalized eigenspace corresponding to the eigenvale zero. the jordan block will look something like $\left( \begin{array}{llll}0 & 1 & 0 &0\cr 0 & 0 & 0 & 0\cr 0 & 0 & 0 & 1\cr 0 & 0 & 0& 0 \end{array} \right).$

similarly you can do the jordan block for the eigenvalue $2.$