## when i use inv to find the inverse matrix of A,i found inv(A)*A is not the identity matrix

### yang-En Hsiao (view profile)

on 14 Nov 2019 at 7:29
Latest activity Answered by Walter Roberson

### Walter Roberson (view profile)

on 14 Nov 2019 at 7:42
Why when i use inv() to find the inverse matrix of matrix H_AB'*H_AB,but their multiplication is not an identity matrix?
H_AB = sqrt(1/2)*[randn(2,7) + j*randn(2,7)];
cc=inv(H_AB'*H_AB)*(H_AB'*H_AB)
The window show me cc is not an identity matrix.So can i still use inv(H_AB'*H_AB) as the identity matrix of H_AB'*H_AB?
Because if the H_AB'*H_AB doesn't have inverser matrix,then how doee matlab calculate?

### Tags ### Walter Roberson (view profile)

on 14 Nov 2019 at 7:42

H_AB = sqrt(1/2)*[randn(2,7) + j*randn(2,7)]
You are building a 2 x 7 matrix. When you then do H_AB'*H_AB then the result has rank 2. inv() cannot be used on rank-deficient matrices.
You could substitute
cc = (H_AB'*H_AB)\(H_AB'*H_AB)
but that will not be an identity matrix either.