In singular value decomposition, where
A = U$\Sigma$V
and A is some n x m matrix, n >= m and rank(A) = m.
can we conclude that the pseudoinverse of $\Sigma$ multiplied by $\Sigma$ is equal to the identity matrix?
My intuition tells me yes, because the pseudoinverse of the singular value(s) matrix is just the recoprical singular values, however I feel like it might not be in the case that A is rectangular, since $\Sigma$ has the same dimensions as A.