# Jordan canonical form with 2 is the dimension of eigenespace,

$$B= \begin{pmatrix} 0 & 1 & 0 \\ -4 & 4 & 0 \\ -2 & 1 & 2 \end{pmatrix}$$ I need to find Jordan decomposition of B.

My sketch:

We find Jordan's matrix first $$A$$ similar to matrix $$B$$, and matrix $$C \in (3,\mathbb{C})$$, such that $$B = C^{-1}AC$$ We have: $$F_{B}(t)=\text{det}(\left( \begin{array}{ccc} -t & 1 & 0 \\ -4 & 4-t & 0 \\ -2 & 1 & 2-t \end{array} \right))=(2-t)(-4t+t^{2}+4)=(2-t)^{3}$$ $$\text{SP}(B) = {2}$$ $$\text{SP}$$ means the spectrum of endomorphism. We are now defining a subspace $$\begin{bmatrix} -2 & 1 & 0 \\ -4 & 2 & 0 \\ -2 & 1 & 0 \end{bmatrix} \begin{bmatrix} x_{1} \\ x_{2} \\ x_{3} \end{bmatrix} \iff x_{1}= -\frac{1}{2}x_{2}$$ So we have two eigenvectors:

$$V_{1}=\begin{bmatrix} -2 \\ 1 \\ 0 \end{bmatrix}$$ $$V_{2}=\begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix}$$ so, $$2$$ is the dimension of the eigenspace, therefore A is not diagonalizable and Jordan canonical form is

$$A=\begin{pmatrix} -2 & 0 & 0 \\ 0 & -2 & 1 \\ 0 & 0 & -2 \end{pmatrix}{}$$

I have issue with last sequence. Why $$B$$ have this specify form $$J$$ with 1 for second row?

• your eigenvalues are 2 not -2 – A. P Aug 18 '19 at 21:12

Firstly since the eigenvalue is 2 we have: $$A=\begin{pmatrix} 2 & 0 & 0 \\ 0 & 2 & 1 \\ 0 & 0 & 2 \end{pmatrix}{}$$ Also by changing the arrangement of the eigenvectors/extend-eigenvectors in the transformation matrix we could also get: $$A=\begin{pmatrix} 2 & 1& 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \end{pmatrix}{}$$ hence the 1 being in the second row is not important or fixed.