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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


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?

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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
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?

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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
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.

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

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