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I have the following inequality to prove.

With $A \in M_n(R)$ show that: $$ (\det(A))^2 \leq \prod_{i=1}^n\left( \sum_{k=1}^n A_{k,i}^2\right) $$

What I already have: I found out that: $$G(v_1,\ldots,v_m) = \det(A^T A)=\det(A^T)\cdot\det(A)=(\det(A))^2 $$

Also that with $G(v_1,\ldots,v_m) = (\det(A))^2$ results that:

$$\operatorname{Vol}(v_1,\ldots,v_m)= \left|\det(A)\right| =\prod_{i=1}^n |s(v_k,u_i)|$$

I don't even now if this is right.. It would be really helpful if someone could help me with the proof..

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There's a quick proof using the $QR$ decomposition. In particular, note that any matrix $A$ can be written as $$ A = QR $$ where $Q$ orthogonal and $R$ is upper triangular (in fact, the columns of $Q$ can be taken as the orthonormal basis attained via the Gram-Schmidt process). Let $q_j$ denote the columns of $Q$, let $a_j$ denote the columns of $j$ (for $j = 1,\dots,n$), and let $r_{ij}$ denote the entries of $R$. Then $$ a_j = \sum_{i=1}^j r_{ij}q_j $$ It follows that $$ \|a_j\|^2 = \sum_{i=1}^j |r_{ij}|^2\|q_j\| \geq |r_{jj}|^2 \implies |r_{jj}| \leq \|a_{j}\| $$ Finally, we have $$ |\det(A)| = |\det(Q)| |\det(R)| = 1 \cdot \left|\prod_{j=1}^n r_{jj}\right| \leq \prod_{j=1}^n \|a_j\| $$ as desired.

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  • $\begingroup$ Thanks! I did read on Wikipedia that you can use QR decomposition, however I was mainly worried because of the (det(A))^2 $\endgroup$ – PhysX May 29 '17 at 18:26
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    $\begingroup$ We could have done it using $A^TA$ directly. In particular, note that $A^TA = R^TR$, and from there we can apply the same inequality to $\det(R^TR) = \det(R)^2$. $\endgroup$ – Omnomnomnom May 29 '17 at 18:31
  • $\begingroup$ Thanks for the information! It's always good to have an alternative. $\endgroup$ – PhysX May 29 '17 at 18:33
  • $\begingroup$ can you use QR if it's not being said that $A$ is a orthogonal matrix? $\endgroup$ – Jneven Jul 16 '18 at 10:07
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    $\begingroup$ @Jneven Yes. Any matrix $A$ will have a $QR$ decomposition, as I state in my answer. Perhaps you'd find the wiki page for QR-decomposition helpful. $\endgroup$ – Omnomnomnom Jul 16 '18 at 10:10

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