Given matrix-valued function A : <mrow class="MJX-TeXAtom-ORD"> <mi mathvaria

Flakqfqbq

Flakqfqbq

Answered question

2022-04-06

Given matrix-valued function A : R d R n × n and (tall) matrix B R n × m , where n > m, let matrix-valued function C : R d R m × m be defined by
C ( x ) := B T A ( x ) B
I would like to maximize the determinant of C ( x ) without directly considering C ( x ). If I maximize the determinant of A ( x ), does that maximize the determinant of C ( x ) as well?

If it were m = n, then we would have:
det ( C ) = det ( B ) 2 det ( A )
Hence, maximising det ( A ) would maximize as a consequence det ( C ). Does it hold something similar in the case with n > m? Or is it possible to define some lower bound for det ( C )? And if so, how to prove it?

If in the previous general case it doesn't hold, maybe it could help that, in my specific case C and A are two positive definite matrices and B is a sparse matrix of zeros and ones. A is also block diagonal. The only way I came up with is using Cauchy-Binet for the determinant of the product of rectangular matrices but I remained stuck with a summation involving the principal minors of the Cholesky factorization of A times minors in B.

Answer & Explanation

parentalite50bqd

parentalite50bqd

Beginner2022-04-07Added 12 answers

Generally speaking, the variable matrix A can vary within some set A . Let the singular value decomposition of B be
B = U D V
hence
C = V T D T U T A U D V
hence
| C | = | D T X D |
where X = U T A U and X belongs to { U T A U | A A }. Now let
D = [ D ^ m × m 0 ( n m ) × m ]
and
X = [ X ( 11 ) ^ m × m X ( 12 ) ^ m × ( n m ) X ( 21 ) ^ ( n m ) × m X ( 22 ) ^ ( n m ) × ( n m ) ]
hence
| C | = | D T X D | = | D ^ T X ( 11 ) D ^ | = | D ^ | 2 | X ( 11 ) | .
Therefore, if B is not full-rank, | C ( x ) | will always be zero, but if it is full-rank, then the maximization of | C ( x ) | is directly the same of | X ( 11 ) | .

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