# Hadamard's inequality proof

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..

• This should be helpful. – Bach Jul 22 at 2:01

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.
• 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 $\det(R^TR) = \det(R)^2$. – Omnomnomnom May 29 '17 at 18:31
• can you use QR if it's not being said that $A$ is a orthogonal matrix? – Jneven Jul 16 '18 at 10:07
• @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. – Omnomnomnom Jul 16 '18 at 10:10
This inequality is 'geometric intuitive' : Note that $$|det(A)|$$ is the volume of the parallelopiped spanned by the $$A_i$$, and that $$\prod_{j = 1}^n ||A_j||_2$$ is the volume of the box with side lengths same lengths as the $$a_i$$. Your assertion is equivalent to the statement that, if you are given fixed side lengths, you can bound the most area when your parallelopiped is a box. This is clear in $$2D$$ by cutting and pasting.
In fact, I think (but haven't checked this carefully) you can prove the statement with such reasoning: If you have any two vectors that are not orthogonal, you can cut and paste to get a new paralellopiped with the same area, where one of the side lengths is strictly smaller. Now you can scale up that length of that side. The end result of doing so results in a paralellopiped with strictly greater volume but with the same side lengths. Keep doing this until all of the vectors are orthogonal, and you are left with a rectangle with volume $$\prod ||A_i||$$, and the course of the proof shows that this is larger than $$|det(A)|$$.