How can I prove the property of ranks for an n x m matrix with n < m?

[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?

 

[chingkui]

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

 

[dabd]

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

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

 

[Office_Shredder]

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.

 

[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?

 

[dabd]

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

 

[JSuarez]

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

 

[dabd]

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

I meant A^T A is symmetric.