Prove A is invertible iff A^T*A is invertible

[physicsss]

Suppose that A is a square. Show that A is invertible if and only if A^T*A is invertible.

I know that if A is an m X n matrix, m>=n, and rkA=n, then the n X n matrix A^T*A is invertible, and that rk(A^TA)= rkA, but I'm still not sure how to start the proof...

TIA

 

[robphy]

Can you use the determinant?

 

[physicsss]

Yes, but how does that help me?

 

[matt grime]

M is invertible if and only if det(M) is not zero.

det(M) is the same as det(M^t)

det(MN)=det(M)det(N)

if x and y are real (or complex) numbers and xy=0 then one of x or y is zero.

 

[Muzza]

det(A) = det(A^t) and det(AB) = det(A)det(B) (for all matrices A, B of the proper size).

If A is invertible, then det(A) != 0, so that det(A^t) != 0, and therefore det(AA^t) = det(A)det(A^t) != 0, i.e. AA^t is invertible.

The converse is similar.

 

[matt grime]

snap. why was the title inner products though?