# Thread: A inequality of determinant

1. ## A inequality of determinant

Assume A is a $n \times m$ real matrix, B is a $n \times (n - m)$ real matrix, show that:
$det\left(\begin{array}{cc}A'A&A'B\\B'A&B'B\end{arr ay}\right) \leq det(A'A)det(B'B)$

2. Originally Posted by Xingyuan
Assume A is a $n \times m$ real matrix, B is a $n \times (n - m)$ real matrix, show that:
$det\left(\begin{array}{cc}A'A&B'A\\A'B&B'B\end{arr ay}\right) \leq det(A'A)det(B'B)$
are you sure you wrote the question correctly?

3. Yes,I am sure.

the left of the equation is not correct, because it is not a square matrix,so the determinant can't define on it .It's my careless,thanks NonCommAlg very much !!!

4. Originally Posted by Xingyuan
Yes,I am sure.
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 $A'A$ and $B'B$ be square matrices because otherwise their determinants (look at the RHS of your inequality) wouldn't be defined. thus $A'$

and $B'$ have to be $m \times n$ and $(n-m)\times n$ respectively. but then $B'A$ would be $(n-m) \times m$ and $A'B$ would be $m \times (n-m).$ 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: $\begin{pmatrix}A'A & A'B \\ B'A & B'B \end{pmatrix}$ ?

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.

5. I am so sorry ,you are right ,the left matrix is $\left(\begin{array}{cc}A'A&A'B\\B'A&B'B\end{array} \right)$.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:

$\left(\begin{array}{cc}A'A&A'B\\B'A&B'B\end{array} \right) = \left(\begin{array}{cc}A'\\B'\end{array}\right)\le ft(\begin{array}{cc}A&B\end{array}\right)$

so we have:

$det\left(\begin{array}{cc}A'A&A'B\\B'A&B'B\end{arr ay}\right) = det(\left(\begin{array}{cc}A'\\B'\end{array}\right )\left(\begin{array}{cc}A&B\end{array}\right))$

by Binet-Cauchy Formular

we have:

$det\left(\begin{array}{cc}A'A&A'B\\B'A&B'B\end{arr ay}\right) = (det\left(\begin{array}{cc}A&B\end{array}\right))^ 2$

so to prove the problem we only need to prove:

$(det\left(\begin{array}{cc}A&B\end{array}\right))^ 2 \leq det(A'A)det(B'B)$

and i can't go any further

another way :

use Laplace theory ,the right of the eqution can be witten as :

$det\left(\begin{array}{cc}A'A&0\\0&B'B\end{array}\ right)$

so if we prove:

$det\left(\begin{array}{cc}A'A&A'B\\B'A&B'B\end{arr ay}\right) \leq det\left(\begin{array}{cc}A'A&0\\0&B'B\end{array}\ right)$

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

6. here's a counter-example for the claim in your problem: let $n = 2, \ m = 1$ and $A=B=\begin{pmatrix}0 \\ 1 \end{pmatrix}, \ A'=\begin{pmatrix}0 & 1 \end{pmatrix}, \ B'= \begin{pmatrix}0 & -1 \end{pmatrix}.$ then $A'A=A'B=1, \ B'A=B'B=-1.$

so we have $\begin{vmatrix} A'A & A'B \\ B'A & B'B \end{vmatrix}=\begin{vmatrix} 1 & 1 \\ -1 & -1 \end{vmatrix}=0> -1=\det(A'A) \det(B'B).$ so there's still something wrong with your problem!

7. If $B = \begin{pmatrix}0 \\ 1 \end{pmatrix}$,then

$B' = \begin{pmatrix}0 & 1 \end{pmatrix}$.
$B' \not=\begin{pmatrix}0 & -1 \end{pmatrix}$

so it is not a contradiction.

8. Originally Posted by Xingyuan
If $B = \begin{pmatrix}0 \\ 1 \end{pmatrix}$,then

$B' = \begin{pmatrix}0 & 1 \end{pmatrix}$.
$B' \not=\begin{pmatrix}0 & -1 \end{pmatrix}$

so it is not a contradiction.
so by $A',B'$ you meant $A^T,B^T$ ?? i thought $A',B'$ were just some other matrices! you see, you should define non-standard notations in your question!

9. Yes , $B'$means $B^T$,the transpose of $B$. In my text book, use this $" ' "$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 ....

10. ok, i haven't finished the proof of the inequality but i think this is the idea that will prove it eventually:

let $\ell=\binom{n}{m}=\binom{n}{n-m}.$ suppose $u_1, \cdots , u_n$ are the rows of $A$ and $v_1, \cdots , v_n$ are the rows of $B.$ by Binet-Cauchy Formula we have $\det(A^TA)=\sum_{i=1}^{\ell} (\det A_i)^2,$ where $A_i$ are all

$m \times m$ minor cofactors of $A.$ call this (1). similarly: $\det(B^TB)=\sum_{i=1}^{\ell} (\det B_i)^2,$ where $B_i$ are all $(n-m) \times (n-m)$ minor cofactors of $B.$ call this (2).

for any $1 \leq i \leq \binom{n}{m},$ we index $A_i$ and $B_i$ somehow that if the rows of $A_i$ are $u_{j_1}, \cdots , u_{j_m},$ then the rows of $B_i$ will be $v_{k_1}, \cdots , v_{k_{n-m}},$ where $\{j_1, \cdots , j_m \} \cap \{k_1, \cdots , k_{n-m} \} = \emptyset.$

thus by (1), (2) and Cauchy-Schwarz inequality we have: $\det(A^TA) \det(B^TB) \geq \left(\sum_{i=1}^{\ell} \det A_i \det B_i \right)^2$ call this (3). clearly we can replace any $\det A_i$ in (1) by $-\det(A_i)$ because

$(\det A_i)^2=(-\det A_i)^2.$ so, we can replace any $\det A_i \det B_i$ in (3) by $-\det A_i \det B_i.$ so, using what you already proved, we only need to prove that for suitable choices of $\pm$ we have:

$\sum_{i=1}^{\ell} \pm \det A_i \det B_i = \det \begin{pmatrix} A & B \end{pmatrix}.$ this identity looks correct to me but i have no proof for it! haha

11. By Laplace Theorem, and pick index $1 < 2 < 3 < \ldots < m$ .
we have:

$\det \begin{pmatrix} A&B \end{pmatrix} = \sum_{1 \leq j_{1} < j_{2} < \ldots < j_{m} \leq n}$ $\det \begin{pmatrix} A&B \end{pmatrix} \begin{pmatrix} j_{1} \ldots j_{m} \\ 1 \ldots m
\end{pmatrix}$
$\delta_{j_{1} j_{2} \ldots j_{n}}^{1 2 \ldots n}$ $\det \begin{pmatrix} A&B \end{pmatrix} \begin{pmatrix} j_{m+1} \ldots j_{n} \\ m+1 \ldots n
\end{pmatrix}$

then for suitable choices of $\pm$, such that equal the correspondence $\delta_{j_{1} j_{2} \ldots j_{n}}^{1 2 \ldots n}$

then we have :

$\sum_{i=1}^{\ell} \pm \det A_i \det B_i = \sum_{1 \leq j_{1} < j_{2} < \ldots < j_{m} \leq n}$ $\det \begin{pmatrix} A&B \end{pmatrix} \begin{pmatrix} j_{1} \ldots j_{m} \\ 1 \ldots m
\end{pmatrix}$
$\delta_{j_{1} j_{2} \ldots j_{n}}^{1 2 \ldots n}$ $\det \begin{pmatrix} A&B \end{pmatrix} \begin{pmatrix} j_{m+1} \ldots j_{n} \\ m+1 \ldots n
\end{pmatrix}$

using what we have already proved, we get the conclusion....

OK, Now I explain the notation that i have used .
the first is:

$\begin{pmatrix} A&B \end{pmatrix} \begin{pmatrix} j_{1} \ldots j_{m} \\ 1 \ldots m
\end{pmatrix}$

this means the sub-matrix of matrix $\begin{pmatrix} A&B \end{pmatrix}$.we got it by choosing the $j_{1},\ldots , j_{m}$ rows, and $1,\ldots ,m$ columns..and the sub-matrix is make up by the cross element of the rows and columns that mentioned above.

the second is :

$\delta_{j_{1} j_{2} \ldots j_{n}}^{1 2 \ldots n}$

the above expression equal $1$ if $j_{1} j_{2} \ldots j_{n}$ through even times transposition become $1 2 \ldots n$.equal $-1$ 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 !!!