"...b must belong to the col(A), which implies rank(A)=m is greater than. or equal to, n"
Let A be an n × m matrix. Prove that if b is a non-zero vector in the nullspace of A transposed, then the systemof linear equations Ax = b has no solution.
So I'm not exactly sure how to go about this.
I know rank(A) = rank(A transposed)
which implies dim(col(A))=dim(col(A transposed))
And for the system to be consistent b must belong to the col(A), which implies rank(A) = n.
But wouldn't the rank(A transposed) = m for the system A(transposed) b = 0 to be consistent which implies rank(A) = m.
But then Ax=b would have no solution right?
if b is a non-zero vector in the null space of AT, then ATb=0, that's what it means to be in the null space.
So now suppose for contradiction that there does exist a solution to Ax=b
then ATAx = ATb = 0
So either x is 0 or ATA is 0. ATA is |A|2 and is only 0 if A=0.
So there are no non-trivial solutions to Ax=b.