well, i'm not sure about that! the reason that asked you that question is that just looking at the dimensions of your matrices shows that the matrix in the LHS of your inequality basically can
never be a square matrix. here's why: we need both and be square matrices because otherwise their determinants (look at the RHS of your inequality) wouldn't be defined. thus
and have to be and respectively. but then would be and would be now see that the matrix you gave us in the LHS just doesn't
make sense! are you sure the matrix in LHS is not this one: ?
by the way your last post (linear algebra) had problems too. next time, maybe nobody will even try to fix your mistakes! so you need to make sure that you've written your question correctly
before submitting it.
I am so sorry ,you are right ,the left matrix is .otherwise the determinant can't define on it ....I am very sorry for my careless,and I promise I will never make these mistakes. and thanks for your help....
My way to solve it(but failed):
because:
so we have:
by Binet-Cauchy Formular
we have:
so to prove the problem we only need to prove:
and i can't go any further
another way :
use Laplace theory ,the right of the eqution can be witten as :
so if we prove:
the problem will be solve.
but i can't do anymore about it.....
At last ,thanks for your help ....and i will never make any mistakes like this...Trust Me
ok, i haven't finished the proof of the inequality but i think this is the idea that will prove it eventually:
let suppose are the rows of and are the rows of by Binet-Cauchy Formula we have where are all
minor cofactors of call this (1). similarly: where are all minor cofactors of call this (2).
for any we index and somehow that if the rows of are then the rows of will be where
thus by (1), (2) and Cauchy-Schwarz inequality we have: call this (3). clearly we can replace any in (1) by because
so, we can replace any in (3) by so, using what you already proved, we only need to prove that for suitable choices of we have:
this identity looks correct to me but i have no proof for it! haha
By Laplace Theorem, and pick index .
we have:
then for suitable choices of , such that equal the correspondence
then we have :
using what we have already proved, we get the conclusion....
OK, Now I explain the notation that i have used .
the first is:
this means the sub-matrix of matrix .we got it by choosing the rows, and columns..and the sub-matrix is make up by the cross element of the rows and columns that mentioned above.
the second is :
the above expression equal if through even times transposition become .equal if through odd times...
hehe, i don't know whether i explain my idea clearly....Hasta La Vista,baby(the problem...)
Thank NonCommAlg very much !!!