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

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yang-En Hsiao
yang-En Hsiao on 14 Nov 2019
Answered: Walter Roberson on 14 Nov 2019
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?

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Answers (1)

Walter Roberson
Walter Roberson on 14 Nov 2019
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.

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