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

Question

yang-En Hsiao 2019-11-14

回答： Walter Roberson 2019-11-14

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?

Answer 1

Walter Roberson 2019-11-14

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.