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I was trying to solve this equation, when i came up with an idea, but couldn't prove it.

The task is: Let the matrices A and B be with the same dimensions. So if A is (2x3) matrice then B is (2x3) matrice. (Is bounds/dimension the right word ? ) Prove that,

$| (A^T * B) | )^2 \le |A^T * A| |B^T * B|$

Anyway i have to prove this , but i thought that determinants can be negative, but i can't find any negative determinants , atleast which look like $|B^T * B|$ . Other thought that if i can show that right side is negative, the equation wouldn't hold , because the right side is always not negative.

My question is : Is there such B, that the Determinant of B transposed multiplied by B is negative ? If not , then why can't it be negative ?

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    $\begingroup$ The question can be found here $\endgroup$
    – user130512
    Feb 26, 2014 at 16:05
  • $\begingroup$ What i posted before was only for square matrices - a proof for the general case can be found above. (to answer your question, then det$(B^tB)$ is simply $det(B)^2$, which can't be negative) $\endgroup$
    – user130512
    Feb 26, 2014 at 16:09
  • $\begingroup$ So even if it is not square matrice, what you said is true ? That proof is too hard for me to understand tho ... It's my 2nd algebra lesson ... haven't gotten to vectors yet. $\endgroup$
    – La'tel
    Feb 26, 2014 at 16:30
  • $\begingroup$ The problem is that the determinant for non-square matrices doesnt exist, it only exists for $B^tB$ because it is square - the link provided is a more general proof for all matrices B $\endgroup$
    – user130512
    Feb 26, 2014 at 16:31
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    $\begingroup$ This is an application of the Binet-Cauchy theorem. In a certain sense, $\det(A^TB)$ is an euclidean scalar product, of outer vector products of the columns of $A$ and $B$ with the coordinate formula $$\det(A^TB)=\sum_{I=(i_1<i_2<...<i_m)}\det(A_I)\,\det(B_I),$$ where $A_I$ is the square matrix with the rows of $A$ indexed by $I$. That is why the formula in the question looks like the Cauchy-Schwarz inequality, since it is a special case of it. But that is not for the second algebra lesson. $\endgroup$
    – Lutz Lehmann
    Feb 26, 2014 at 19:30

1 Answer 1

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To be non-trivial, the format of the matrices must be "tall", more rows than columns, $m\ge n$. (If not, $\det(A^TA)=\det(A^TB)=\det(B^TB)=0$ because of non-maximal rank of the products.)

Now use QR decompositions $A=UR$ and $B=VS$ in the "small" variant where $U$ and $V$ have the same format as $A$ and $B$ with orthonormal columns. $R$ and $S$ are upper triangular $n\times n$ matrices. Then \begin{align} \det(A^TA)&=\det(R^TR)=\det(R)^2\\ \det(B^TB)&=\det(S^TS)=\det(S)^2\\ \det(A^TB)&=\det(R^T(U^TV)S)=\det(R)\det(S)\det(U^TV) \end{align} so the problem reduces to the special case $|\det(U^TV)|\le 1$. Here one can perhaps use the spat volume property of the determinant, and its maximum for given side lengths, $$ |\det(\vec x_1,\vec x_2,...,\vec x_n)|\le \|\vec x_1\|\,\|\vec x_2\|\,...\,\|\vec x_n\| $$ that is, applied to the current situation, $$ |\det(U^TV)|\le \|U^Tv_1\|\,\|U^Tv_2\|\,...\,\|U^Tv_n\|. $$

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