Consider the m &#x00D7;<!-- × --> m -matrix B , which is symmetric and positive defi



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Consider the m × m-matrix B, which is symmetric and positive definite (full rank). Now this matrix is transformed using another matrix, say A, in the following manner: A B A T . The matrix A is n × m with n < m. Furthermore the constraint r a n k ( A ) < n is imposed.
My intuition tells me that A B A T must be symmetric and positive semi-definite, but what is the mathematical proof for this? (why exactly does the transformation preserve symmetry and why is it that possibly negative eigenvalues in A still result in the transformation to be PSD? Or is my intuition wrong)?

Answer & Explanation

Kiana Cantu

Kiana Cantu

Beginner2022-07-04Added 22 answers

For symmetry: note that in general, we have ( A B ) T = B T A T , hence ( A B C ) T = C T B T A T . With that, we see that
( A B A T ) T = A T T B T A T = A B A T .
For positive semidefiniteness: an n × n symmetric matrix M is positive semidefinite if (and only if) x T M x 0 whenever x R n . We note that
x T ( A B A T ) x = ( x T A ) B ( A T x ) = ( A T x ) T B ( A T x ) .
Because B is positive definite (and hence positive semidefinite), we must have y T B y 0 for y = A T x. Thus, x T ( A B A T ) x 0, so that A B A T is indeed positive semidefinite.

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