Thread: matrix and determinants

Let A be an m x n matrix, where m<n, then A^t(A) is an n by n matrix. show that det(A^t(A))=0

also if A is a 2 x 3 matrix can you say that det (AA^t)=0

thanks

Originally Posted by b0mb3rz

Let A be an m x n matrix, where m<n, then A^t(A) is an n by n matrix. show that det(A^t(A))=0

also if A is a 2 x 3 matrix can you say that det (AA^t)=0

thanks

I'm not the best at Linear Algebra but I'll give this a crack until someone else helps you out.

I'm assuming A^t(A) to mean the transpose matrix of A. So if A is an m x n matrix then the transpose should be n x m, since the rows and columns switch.

Before I try to help, is this what you meant?

yeah it is and it says hint, consider the system of homogeneous linear equations Ax=0

Originally Posted by b0mb3rz

yeah it is and it says hint, consider the system of homogeneous linear equations Ax=0

Well determinants only apply to square matrices, so I don't see how this problem is valid.

Originally Posted by Jameson

Well determinants only apply to square matrices, so I don't see how this problem is valid.

b0mb3rz didn't probably make it clear but i think what he meant was this is not difficult to prove:

thus, by the rank-nullity theorem, the null-space of is non-zero and hence


you may also argue that: