Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I have to show that each of the following matrices

$$ \frac{1}{\sqrt{2}} \begin{pmatrix} 0&1&0\\ 1&0&1\\ 0&1&0 \end{pmatrix}\quad , \frac{1}{\sqrt{2}} \begin{pmatrix} 0&-i&0\\ i&0&-i\\ 0&i&0 \end{pmatrix} , \begin{pmatrix} 1&0&0\\ 0&0&0\\ 0&0&-1 \end{pmatrix}$$

are equivalent to one of the following

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

It is a relatively simple but time consuming problem to construct a similarity transformation. I have nine real independent variables and 9 equations to check for each matrix. I want to know if there is any process or software that could help me do this calculation?

share|improve this question
1  
Which kind of "equivalent" are you using? –  Henning Makholm Dec 8 '11 at 23:24
    
@Henning Sorry. I am following a physics book and these matrices are actually representations. Two representations were defined to be equivalent if their matrices are related by a similarity transformation. This was what I had in my mind. –  kuch nahi Dec 9 '11 at 0:36

2 Answers 2

up vote 1 down vote accepted

Since you mentioned a "similarity transform", I presume your "equivalent" means "similar"; also I presume this is over $\mathbb C$. Well, the first thing I would do is find the eigenvalues of each matrix; the second (if necessary) is the Jordan canonical form. As for software, Maple will handle this quite easily; I imagine most other CAS's will also.

EDIT: Oops, all of your first three matrices are similar to each other, and none is similar to any of the last three. Are you sure you quoted the question right?

share|improve this answer
    
Yes, I am sure. The first three matrices are what physicists call spin 1 representation. The final three matrices are the adjoint representation of SU(2) (excluding a phase factor of -i). The question is to show that the two representations are equivalent (a result which is used later to construct roots). I phrased only the computational part in my question. –  kuch nahi Dec 9 '11 at 0:28
    
@kuchnahi The top matrices are Hermitian hence their eigenvalues are real. Bottom left two are skew-symmetric and hence have eigenvalues pure imaginary. This means they can't be similar. The only option is the bottom right but that also has complex eigenvalues. –  user13838 Dec 9 '11 at 1:26
    
The bottom right doesn't work already because it is invertible and none of the other 5 are. –  Henning Makholm Dec 9 '11 at 12:27
    
Sorry,you all were right. I made a type in the third matrix –  kuch nahi Dec 9 '11 at 15:50
1  
In fact, if the top 3 matrices are $A_j$, consider $U A_j U^{-1}$ where $U = \pmatrix{-1/\sqrt{2} & 0 & 1/\sqrt{2}\cr -i/\sqrt{2} & 0 & -i/\sqrt{2}\cr 0 & 1 & 0\cr}$ –  Robert Israel Dec 9 '11 at 18:10

Two matrices are similar if their traces are equal.

share|improve this answer
    
So every traceless matrix is the zero matrix? –  EuYu Nov 4 '12 at 4:21
4  
This is true only for $1\times 1$ matrices. –  Jason DeVito Nov 4 '12 at 4:22

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.