Given matrix-valued function A : R d → R n × n and (tall) matrix B ∈ R n × m , where n &gt; m, let matrix-valued function C : R d → R m × m be defined by C ( x ) := B T A ( x ) BI 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 &gt; 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.

Question

Given matrix-valued function   A  :            R        d    →            R              n      ×      n       and (tall) matrix   B  ∈            R              n      ×      m      , where   n  &amp;gt;  m, let matrix-valued function   C  :            R        d    →            R              m      ×      m       be defined by  C  (  x  )  :=      B    T    A  (  x  )  BI 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  &amp;gt;  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&#039;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.

parentalite50bqd · Accepted Answer

Generally speaking, the variable matrix   A can vary within some set       A  . Let the singular value decomposition of   B be  B  =  U  D  Vhence  C  =      V    T        D    T        U    T    A  U  D  Vhence      |    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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