# Prove property of ranks

1. Jun 8, 2010

### dabd

if A is an n x m matrix where n < m I would like to prove that there exists some $$\lambda$$ such that $$rank(A^T A + \lambda I) = m$$

I know that if two of the columns of $$A^T A$$ are linearly dependent, they are scalar multiples of each other and by adding some $$\lambda$$ to two different positions, those colums will become independent but I can't prove it for more than two columns.

Any tips?

Last edited: Jun 8, 2010
2. Jun 8, 2010

### chingkui

How many eigen-values does transpose(A)*A has?

3. Jun 8, 2010

### dabd

That is not given. A is any n x m matrix.

4. Jun 8, 2010

### Office_Shredder

Staff Emeritus
It seems to me that you're taking the pessimistic viewpoint. The rank will be full if the determinant is non-zero. The determinant of $$A^T A + \lambda I$$ is a polynomial in $$\lambda$$ , so will have non-zero determinant for all but finitely many $$\lambda$$.

5. Jun 8, 2010

### JSuarez

A few hints:

(1) $A^T A + \lambda I$ is full rank iff it's invertible.

(2) Any strictly Diagonally dominant matrix is invertible.

What are the diagonal elements of $A^T A + \lambda I$?

6. Jun 8, 2010

### dabd

I got the answer in simpler terms $$A^T A$$ is symmetric, so it is positive semi-definite and by taking any $$\lambda > 0$$ the matrix $$A^T A + \lambda I$$ is positive definite, hence non-singular and invertible.

Last edited: Jun 9, 2010
7. Jun 8, 2010

### JSuarez

If B is symmetric, then -B is positive-semidefinite? Notwithstanding this, you can choose $\lambda$ such that your matrix is indeed positive definite.

8. Jun 9, 2010

### dabd

I meant $$A^T A$$ is symmetric.