# Relation between rank and number of non-zero eigen values.

1. Dec 26, 2008

### NaturePaper

Hi everyone,
I've a simple question (the answer may be so trivial that I really ought to be ashamed for asking!) in elementary matrix theory:
"Does there exists any relation between the number of non-zero eigen values of a matrix with its rank?" The matrix is taken to be a general (square, of course) matrix with complex entries.

[some partial result for full-ranked matrix is known to me, but I want the general relation, if it exists]

What if we restricted to Hermitian (and more specially to Positive semidefinite) matrices?

Thanks & Regards,
Naturepaper

2. Dec 26, 2008

### HallsofIvy

Staff Emeritus
There is a very fundamental theorem that says if L is a linear transformation from Rn to Rm, then the rank of L (dimension of L(Rn) plus the nullity of L (dimension of kernel of L) equals m.

In order to talk about the eigenvalues of a matrix, it must be from Rn to Rn, square as you say: the rank plus nullity = n. If v is in the nullity of L then Lv= 0 so v is an eigenvector with eigenvalue 0. Since there exist an "eigenspace" of degree at least one for each eigenvalue, the number of non-zero eigenvalues must be less than or equal to the rank. More than that you can't say.

3. Dec 26, 2008

### NaturePaper

Perhaps the theorem you mentioned is well-known as :'Sylvaster law of nullity' (forgive the spelling mistakes). I'm talking about matrices with complex entries.

4. Dec 26, 2008

### NaturePaper

Just seen a relation, somehow interesting: 'rank=number of non-zero singular values'

[p. 31 of 'Eigenvalues of Matrices' by Francoise Chatelin]

5. Dec 27, 2008

### HallsofIvy

Staff Emeritus
If you are counting multiplicity, of course. The characteristic equation of an n by n matrix is an n-degree polynomial and so has n (not all different) complex roots. If the nullity is k, then k of the number of eigenvalues equal to 0 so there are n- k non-zero roots left. By the theorem I quoted above, that us the rank.

6. Dec 28, 2008

### NaturePaper

"Number of non-zero eigenvalues<= rank" for all complex square matrices.

7. Dec 29, 2008

### HallsofIvy

Staff Emeritus
I had already written "yes" (as I wrote "of course" before!) when I decided to stop and rethink what I had said. My statement before, "If the nullity is k, then k of the number of eigenvalues equal to 0" is not true! For example, the matrix
$$\left[\begin{array}{cc}0 & 1 \\ 0 & 0\end{array}\right]$$
obviously has 0 as a double eigenvalue and has no non-zero eigenvalues. But just as obviously the null space is all vectors of the form <x, 0> which is one-dimensional and the range is all vectors of the form <0, y>, again one dimensional. Now, I know perfectly well that the dimension of an eigenspace may be lower that the multiplicity of the eigenvalue. I don't know why I assumed that must be the case for eigenvalue 0! Sorry about that. I think the best that can be said is that if an n by n matrix has at least one eigenvalue then the rank must be less than n. To extend the example above, the matrix having "1"s along the diagonal just above the main diagonal and "0"s everywhere else, has all n eigenvalues equal to 0 but has rank n-1.

8. Dec 29, 2008

### NaturePaper

Oh sure, I've noted that previous.

In addition, in your first post, it was better to write V^n instaed of R^n and to write "range+nullity=dim V for finite dimensional vector space V" instead of "There is a very fundamental theorem that says if L is a linear transformation from Rn to Rm, then the rank of L (dimension of L(Rn) plus the nullity of L (dimension of kernel of L) equals m." Because I've asked for complex matrices so it is from C^n to C^n not from R^n to R^n.

Regarding your last post: "I think the best that can be said is that if an n by n matrix has at least one eigenvalue then the rank must be less than n", is also not true as can be verified for the (simply) I_n, the identity matrix of order n (in fact any non-singular matrix)!

(I think by the phrase "has at least one eigenvalue" you meant "has at least one non-zero eigenvalue")

Regards,
Naturepaper

Last edited: Dec 29, 2008
9. Dec 29, 2008

### NaturePaper

Again, surprisingly I've got (possibly) the answer for the last part of my question:

For NxN hermitian matrix, "rank=number of non-zero eigen values"!!!

I don't know how to prove it but here is the reference where I got it : arXiv:quant-ph/0403151v1
(see 3rd line of page no. 10)

10. Dec 29, 2008

### Marcaias

Suppose A is n-by-n, hermitian, and exactly k of its eigenvalues are 0. Because it is hermitian, it is diagonalizable and all of its eigenvalues are real-valued. Therefore, in some basis A is equal (similar) to a real-valued, diagonal matrix where k of the diagonal terms are 0, and n-k are nonzero. The rank of this matrix is clearly n-k.

11. Dec 29, 2008

### Marcaias

By the way, you don't even need the hermitian property. The same argument works for any diagonalizable matrix A.

Last edited: Dec 29, 2008
12. Dec 29, 2008

Oh, nice.

Thanks.