# How can I make singular matrix become nonsingular matrix?

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1. Oct 15, 2016

### munirah

<< Mentor Note -- thread moved from Homework Help forums to General Math >>

Good day,

I run coding in Mathematica. But, I get singular matrix A at certain loop. In theory, how can I make matrix A become orthogonal

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

because I want to make a unitary transformation on matrix B

B=\begin{pmatrix} 1& 0 &
0 & 0 & 0 & 0 & 0 & 0\\ 0& 0 &
0 & 0 & 0 & 0 & 0 & 0 \\ 0& 0 &
0 & 0 & 0 & 0 & 0 & 0\\ 0& 0 &
0 & 0 & 0 & 0 & 0 & 0\\ 0& 0 &
0 & 0 & 0 & 0 & 0 & 0\\ 0& 0 &
0 & 0 & 0 & 0 & 0 & 0
\\ 0& 0 &
0 & 0 & 0 & 0 & 0 & 0\\ 0& 0 &
0 & 0 & 0 & 0 & 0 & 0
\end{pmatrix}

with using definition AB A^{-1} since matrix A is orthogonal matrix.

Thank you very much.

Last edited by a moderator: Oct 15, 2016
2. Oct 16, 2016

### strangerep

Afaict, you haven't given us enough information to enable us to help you.

Your $A$ matrix is singular. Without further information, you can't magically make it nonsingular.

3. Oct 16, 2016

### munirah

Thank you for respond. I want to make unitary transformation to the state B. When I run the coding, at certain time, the matrix become singular. How can I solve it hence it will give the result of the trace equal to 1

4. Oct 16, 2016

### strangerep

Sorry, but unless you can better understand the meaning of "you haven't given us enough information", I doubt anyone can help you.

5. Oct 16, 2016

### Staff: Mentor

You could vary the matrix entries by a small (infinitesimal) amount to get it regular or even unitary. But as long as you cannot control the meaning of such a manipulation, your results become worthless. It would be a topological method, and you have to assure beforehand, that this is a valid method to do. Normally, i.e. when considered the pure algebraic or geometrical aspects of a linear equation, this won't be allowed, or to be exact: only within very special conditions. (V. Strassen once considered aspects of the Zariski-topology to prove lower bounds on the complexity of matrix multiplication.)

This means you can't do this, e.g. to get rid of the kernel.
So in the generality you stated your question, the answer has very likely to be: It cannot be done (without smashing the outcome).

6. Oct 16, 2016

### FactChecker

Your A matrix is so far from orthogonal that it just zeros out 6 of the 8 dimensions. And it zeros out the only dimension where B has a non-zero element. So making A orthogonal would be totally fabricated.

That being said, I will take a wild stab at something. If you are asking for an orthogonal matrix that has the -1s of A but leaves the other dimensions unchanged (instead of zeroed out), then this might be what you want. It just negates and swaps the 3rd and 5th dimensions and it negates and swaps the 4th and 6th dimensions, leaving the other dimensions 1,2,7,8 unchanged.

C =\begin{pmatrix}
1& 0 & 0 & 0 & 0 & 0 & 0 & 0\\
0& 1 & 0 & 0 & 0 & 0 & 0 & 0 \\
0& 0 & 0 & 0 & -1 & 0 & 0 & 0 \\
0& 0 & 0 & 0 & 0 & -1 & 0 & 0\\
0& 0 & -1 & 0 & 0 & 0 & 0 & 0\\
0& 0 & 0 & -1 & 0 & 0 & 0 & 0\\
0& 0 & 0 & 0 & 0 & 0 & 1 & 0\\
0& 0 & 0 & 0 & 0 & 0 & 0 & 1
\end{pmatrix}

Last edited: Oct 16, 2016
7. Oct 17, 2016

### munirah

Thank you for respond it. It check it and yes,I made a mistake from multiply the matrix. Thank you all.