Assume A is a $\displaystyle n \times m$ real matrix, B is a $\displaystyle n \times (n - m)$ real matrix, show that:
$\displaystyle det\left(\begin{array}{cc}A'A&A'B\\B'A&B'B\end{arr ay}\right) \leq det(A'A)det(B'B)$
Assume A is a $\displaystyle n \times m$ real matrix, B is a $\displaystyle n \times (n - m)$ real matrix, show that:
$\displaystyle det\left(\begin{array}{cc}A'A&A'B\\B'A&B'B\end{arr ay}\right) \leq det(A'A)det(B'B)$
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 $\displaystyle A'A$ and $\displaystyle B'B$ be square matrices because otherwise their determinants (look at the RHS of your inequality) wouldn't be defined. thus $\displaystyle A'$
and $\displaystyle B'$ have to be $\displaystyle m \times n$ and $\displaystyle (n-m)\times n$ respectively. but then $\displaystyle B'A$ would be $\displaystyle (n-m) \times m$ and $\displaystyle A'B$ would be $\displaystyle m \times (n-m).$ 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: $\displaystyle \begin{pmatrix}A'A & A'B \\ B'A & B'B \end{pmatrix}$ ?
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 $\displaystyle \left(\begin{array}{cc}A'A&A'B\\B'A&B'B\end{array} \right)$.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:
$\displaystyle \left(\begin{array}{cc}A'A&A'B\\B'A&B'B\end{array} \right) = \left(\begin{array}{cc}A'\\B'\end{array}\right)\le ft(\begin{array}{cc}A&B\end{array}\right)$
so we have:
$\displaystyle det\left(\begin{array}{cc}A'A&A'B\\B'A&B'B\end{arr ay}\right) = det(\left(\begin{array}{cc}A'\\B'\end{array}\right )\left(\begin{array}{cc}A&B\end{array}\right))$
by Binet-Cauchy Formular
we have:
$\displaystyle det\left(\begin{array}{cc}A'A&A'B\\B'A&B'B\end{arr ay}\right) = (det\left(\begin{array}{cc}A&B\end{array}\right))^ 2$
so to prove the problem we only need to prove:
$\displaystyle (det\left(\begin{array}{cc}A&B\end{array}\right))^ 2 \leq det(A'A)det(B'B)$
and i can't go any further
another way :
use Laplace theory ,the right of the eqution can be witten as :
$\displaystyle det\left(\begin{array}{cc}A'A&0\\0&B'B\end{array}\ right)$
so if we prove:
$\displaystyle det\left(\begin{array}{cc}A'A&A'B\\B'A&B'B\end{arr ay}\right) \leq det\left(\begin{array}{cc}A'A&0\\0&B'B\end{array}\ right)$
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
here's a counter-example for the claim in your problem: let $\displaystyle n = 2, \ m = 1$ and $\displaystyle A=B=\begin{pmatrix}0 \\ 1 \end{pmatrix}, \ A'=\begin{pmatrix}0 & 1 \end{pmatrix}, \ B'= \begin{pmatrix}0 & -1 \end{pmatrix}.$ then $\displaystyle A'A=A'B=1, \ B'A=B'B=-1.$
so we have $\displaystyle \begin{vmatrix} A'A & A'B \\ B'A & B'B \end{vmatrix}=\begin{vmatrix} 1 & 1 \\ -1 & -1 \end{vmatrix}=0> -1=\det(A'A) \det(B'B).$ so there's still something wrong with your problem!
Yes ,$\displaystyle B'$means $\displaystyle B^T$,the transpose of $\displaystyle B$. In my text book, use this $\displaystyle " ' "$notation to show a transpose of a matrix....next time ,if i use any notation in my problem. i will explain it clearly... 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 $\displaystyle \ell=\binom{n}{m}=\binom{n}{n-m}.$ suppose $\displaystyle u_1, \cdots , u_n$ are the rows of $\displaystyle A$ and $\displaystyle v_1, \cdots , v_n$ are the rows of $\displaystyle B.$ by Binet-Cauchy Formula we have $\displaystyle \det(A^TA)=\sum_{i=1}^{\ell} (\det A_i)^2,$ where $\displaystyle A_i$ are all
$\displaystyle m \times m$ minor cofactors of $\displaystyle A.$ call this (1). similarly: $\displaystyle \det(B^TB)=\sum_{i=1}^{\ell} (\det B_i)^2,$ where $\displaystyle B_i$ are all $\displaystyle (n-m) \times (n-m)$ minor cofactors of $\displaystyle B.$ call this (2).
for any $\displaystyle 1 \leq i \leq \binom{n}{m},$ we index $\displaystyle A_i$ and $\displaystyle B_i$ somehow that if the rows of $\displaystyle A_i$ are $\displaystyle u_{j_1}, \cdots , u_{j_m},$ then the rows of $\displaystyle B_i$ will be $\displaystyle v_{k_1}, \cdots , v_{k_{n-m}},$ where $\displaystyle \{j_1, \cdots , j_m \} \cap \{k_1, \cdots , k_{n-m} \} = \emptyset.$
thus by (1), (2) and Cauchy-Schwarz inequality we have: $\displaystyle \det(A^TA) \det(B^TB) \geq \left(\sum_{i=1}^{\ell} \det A_i \det B_i \right)^2$ call this (3). clearly we can replace any $\displaystyle \det A_i$ in (1) by $\displaystyle -\det(A_i)$ because
$\displaystyle (\det A_i)^2=(-\det A_i)^2.$ so, we can replace any $\displaystyle \det A_i \det B_i$ in (3) by $\displaystyle -\det A_i \det B_i.$ so, using what you already proved, we only need to prove that for suitable choices of $\displaystyle \pm$ we have:
$\displaystyle \sum_{i=1}^{\ell} \pm \det A_i \det B_i = \det \begin{pmatrix} A & B \end{pmatrix}.$ this identity looks correct to me but i have no proof for it! haha
By Laplace Theorem, and pick index $\displaystyle 1 < 2 < 3 < \ldots < m $ .
we have:
$\displaystyle \det \begin{pmatrix} A&B \end{pmatrix} = \sum_{1 \leq j_{1} < j_{2} < \ldots < j_{m} \leq n}$$\displaystyle \det \begin{pmatrix} A&B \end{pmatrix} \begin{pmatrix} j_{1} \ldots j_{m} \\ 1 \ldots m
\end{pmatrix}$$\displaystyle \delta_{j_{1} j_{2} \ldots j_{n}}^{1 2 \ldots n}$$\displaystyle \det \begin{pmatrix} A&B \end{pmatrix} \begin{pmatrix} j_{m+1} \ldots j_{n} \\ m+1 \ldots n
\end{pmatrix}$
then for suitable choices of $\displaystyle \pm$, such that equal the correspondence $\displaystyle \delta_{j_{1} j_{2} \ldots j_{n}}^{1 2 \ldots n}$
then we have :
$\displaystyle \sum_{i=1}^{\ell} \pm \det A_i \det B_i = \sum_{1 \leq j_{1} < j_{2} < \ldots < j_{m} \leq n}$$\displaystyle \det \begin{pmatrix} A&B \end{pmatrix} \begin{pmatrix} j_{1} \ldots j_{m} \\ 1 \ldots m
\end{pmatrix}$$\displaystyle \delta_{j_{1} j_{2} \ldots j_{n}}^{1 2 \ldots n}$$\displaystyle \det \begin{pmatrix} A&B \end{pmatrix} \begin{pmatrix} j_{m+1} \ldots j_{n} \\ m+1 \ldots n
\end{pmatrix}$
using what we have already proved, we get the conclusion....
OK, Now I explain the notation that i have used .
the first is:
$\displaystyle \begin{pmatrix} A&B \end{pmatrix} \begin{pmatrix} j_{1} \ldots j_{m} \\ 1 \ldots m
\end{pmatrix}$
this means the sub-matrix of matrix$\displaystyle \begin{pmatrix} A&B \end{pmatrix}$.we got it by choosing the $\displaystyle j_{1},\ldots , j_{m} $ rows, and $\displaystyle 1,\ldots ,m$ columns..and the sub-matrix is make up by the cross element of the rows and columns that mentioned above.
the second is :
$\displaystyle \delta_{j_{1} j_{2} \ldots j_{n}}^{1 2 \ldots n}$
the above expression equal $\displaystyle 1$ if $\displaystyle j_{1} j_{2} \ldots j_{n}$ through even times transposition become $\displaystyle 1 2 \ldots n$.equal $\displaystyle -1$ 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 !!!