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Compute the requirement for the rank of a matrix?

Posted 7 years ago

Hey There!

I have a 8x8 matrix with four variables in it. I want to compute the solution for the variables where the rank of the matrix gets four. I know it has something to do with the eigenvalues but not exactly how to solve it.

Thanks for your help!

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3 Replies

The eigenvalues are the roots of the characteristic polynomial. Factored, that polynomial is (as a poly in x):

-(2 qd + 2 qv - x) x (4 qd qo + 4 qd qu + 16 qo qu + 4 qd qv + 
   4 qo qv + 4 qu qv - 2 qd x - 4 qo x - 4 qu x - 2 qv x + 
   x^2) (2 qd qo + 2 qd qu + 4 qo qu + 2 qd qv + 2 qo qv + 2 qu qv - 
   2 qd x - 2 qo x - 2 qu x - 2 qv x + x^2)^2

The problem at hand is to find relations amongst the matrix elements so that exactly four eigenvalues are zero. One is zero independent of the matrix element values. If you zero the quadratic of multiplicity 2 then the rank is 3, so the only option is to zero the remaining factors.

   In[11]:= solns = 
 Solve[{{(2 qd + 2 qv - x), (4 qd qo + 4 qd qu + 16 qo qu + 4 qd qv + 
        4 qo qv + 4 qu qv - 2 qd x - 4 qo x - 4 qu x - 2 qv x + 
        x^2)} == 
     0, (2 qd qo + 2 qd qu + 4 qo qu + 2 qd qv + 2 qo qv + 2 qu qv - 
       2 qd x - 2 qo x - 2 qu x - 2 qv x + x^2) != 0} /. x -> 0]

(* Out[11]= {{qu -> qd^2/(4 qo), qv -> -qd}} *)
POSTED BY: Daniel Lichtblau

And I assume that there are several possibilities to get rank = 4 for your matrix. See attached notebook

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POSTED BY: Hans Dolhaine

I think the rank of a matrix is equal to the number of nonzero-eigenvalues. In the the notebook attached I tried to find combinations of the q's in order to generate zero-eigenvalues, assuming that you are interested in matrix De. Well, I arrived at four eigenvalues == 0 , so with this combinations the rank of De is 4.

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POSTED BY: Hans Dolhaine
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