Finding the Jordan decomposition

  • #1
274
1

Homework Statement


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}\}##.


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.
 

Answers and Replies

  • #2
vela
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It's okay to have zeros on the diagonal.
 
  • #3
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It's okay to have zeros on the diagonal.
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
vela
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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
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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?
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
vela
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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
vela
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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.
It seems to go by the name Jordan-Chevalley decomposition.
 

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