Relationship between eigenvalues and matrix rank

[Fluger]

I'm looking into the stability of a system of ODEs, for which we've mannaged to extract a Jacobian matrix. Two of our eigenvalues are within our nummerical error tolerance, but they are close to zero. One of them is positive, which poses a problem for our stability analysis.

We do know that the rank of the matrix is 17, against the 19 variables we are studying (19x19 matrix). I'm guessing that this might imply that our two eigenvalues are in fact zeroes, but I'm having trouble putting anything more concrete down to paper. Do you guys know of any relationship between how many eigenvalues there are in zero and the nullity of the matrix?

 

[weetabixharry]

If the rank of the matrix is 17, then there are exactly 17 non-zero eigenvalues.

If you're dealing with numerical values on a (finite-precision) computer, then the 'zero' eigenvalues may be different from zero by some small amount.

 

[DonAntonio]

If the rank of the matrix is 17, then there are exactly 17 non-zero eigenvalues.

If you're dealing with numerical values on a (finite-precision) computer, then the 'zero' eigenvalues may be different from zero by some small amount.

The above isn't accurate. The matrix [tex]\left(\begin{array}{ccc} 0&1&1\\0&0&1\\0&0&2\end{array}\right)[/tex] has rank 2 but one single non-zero eigenvalue.

DonAntonio

 

[weetabixharry]

Sorry - for some reason I had thought our matrix was Hermitian. If the matrix isn't diagonalisable, please ignore what I wrote.

 

[Citan Uzuki]

The nullity of the matrix is the geometric multiplicity of the eigenvalue zero. In general, the geometric multiplicity of an eigenvalue is less than or equal to its algebraic multiplicity. So if your matrix has rank 17, then at least 2 of your eigenvalues are zero. If you've excluded the other 17, then that means the two that are within your error tolerance must both be zero.