# Finding the Jordan decomposition

1. Oct 6, 2013

1. The problem statement, all variables and given/known data
I have a linear transformation T defined by
$$T(v_{1})=v_{1}+iv_{2}\\ T(v_{2})=-iv_{1}+v_{2}\\$$
and I want to find a triangular matrix B of T and an invertible matrix S such that SB=AS where A is the matrix of T with respect to the basis $\{v_{1},v_{2}\}$.

3. The attempt at a solution
The matrix A is
$$\begin{pmatrix} 1 & -i \\ i & 1 \\ \end{pmatrix}.$$
From the proof of the triangular form theorem, I know that I need to find a new basis $\{w_{1},w_{2}\}$ such that $Tw_{1}=0$ and $(T-2)w_{2}=0$ since the minimal polynomial is $m(T)=T(T-2)=0$ but I must be messing something up because there is no way I can create a triangular matrix B from $w_{1}$ and $w_{2}$. If I find a linear combination of $v_{1},v_{2}$ equal to $w_{1}$ such that $Tw_{1}=0$ then the first column of my matrix B would always be 0.

2. Oct 6, 2013

### vela

Staff Emeritus
It's okay to have zeros on the diagonal.

3. Oct 6, 2013

Hmm. I am choosing the new basis $w_{1}=v_{1}-iv_{2}$ and $w_{2}=v_{1}+iv_{2}$ which gives me
$$B=\begin{pmatrix} 0 & 0 \\ 0 & 2 \\ \end{pmatrix}$$
and
$$S=\begin{pmatrix} 1 & 1 \\ -i & i \\ \end{pmatrix}$$
which gives me the SB=AS that I need but then when I do the Jordan decomposition is it okay to have
$$D=\begin{pmatrix} 0 & 0 \\ 0 & 2 \\ \end{pmatrix}$$
and
$$N=\begin{pmatrix} 0 & 0 \\ 0 & 0 \\ \end{pmatrix}$$
so that T=D+N?

4. Oct 6, 2013

### vela

Staff Emeritus
I'm afraid I'm not familiar with what you're trying to do. D, I assume, is supposed to be a diagonal matrix. What is N defined to be?

5. Oct 6, 2013

I am trying to find the Jordan Decomposition of T. That is what it is called in my book. From a quick search it doesn't look like it goes by the name of Jordan Decomposition on wikipedia.

In my book D is defined to be a diagonalizable linear transformation while N is defined to be a nilpotent linear transformation such that T=D+N.

6. Oct 6, 2013

### vela

Staff Emeritus
Your N is definitely nilpotent, so that decomposition looks fine to me. Since the transformation you were given doesn't have degenerate eigenvalues, the associated eigenvectors will form a basis, and you should expect the similarity transformation to yield a diagonal matrix and to therefore have N=0.

If you had degenerate eigenvalues, it's then possible that the eigenvectors won't form a basis, and you'd have to resort to generalized eigenvectors. In this case, the similarity transformation would yield an almost-diagonal matrix, the so-called Jordan normal form, and N would have non-zero entries just above some of the diagonal elements.

7. Oct 6, 2013

### vela

Staff Emeritus
It seems to go by the name Jordan-Chevalley decomposition.