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I've asked a similar problem earlier here and I was told the same is true for $\lvert\det(A)| \leq \prod_{i=1}^n \Vert a_{i}\Vert$

with $\Vert a_{i,i}\Vert$ being the norm induced by the Euclidean inner product so that $\Vert a_{i}\Vert = \sqrt {a_i \overline{a_i}}$

Moreover, $a_i$ is the i-th column of the pos. def. Matrix $A \in \mathbb{C^{n\times n}}$.

Like in my other question I've tried to work with the Cholesky-decomposition because $\prod_{i=1}^n \Vert a_{i}\Vert$ is basically the same operation as $\prod_{i=1}^n a_{i,i}$ with the difference, that the first product square-roots all elements.

Sadly, my attempt failed because it seems like $\det(A)$ is actually greater than the product.

Can anyone help me finding a good proof?

John B
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Quotenbanane
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1 Answers1

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It is a bit more delicate. On the other hand more is true.

For details you may have a look at:

S. Afriat, Orthogonal and oblique projectors and the characteristics of pairs of vector spaces, Proc. Camb. Philos. Soc. 53 (1957), 800-816.

I will sum it up (since it would be impossible to give all the details):

The volume ${\rm vol}(a_1,\ldots,a_n)$ determined by the $n$ vectors $a_1,\ldots,a_n$ on the columns of $A$ satisfies $$ {\rm vol}(a_1,\ldots,a_n)=\|a_1\|{\rm vol}(a_2,\ldots,a_n)\sin\theta_1, $$ where $\theta_1$ is the angle between $a_1$ and the $(n-1)$-dimensional space determined by $a_2,\ldots,a_n$ (details in the paper above, too lengthy for a reply). Now you iterate this procedure to obtain $$ \lvert\det A\rvert={\rm vol}(a_1,\ldots,a_n)=\prod_{i=1}^n\|a_i\|\prod_{i=1}^{n-1}\sin\theta_i\le\prod_{i=1}^n\|a_i\|, $$ where $\theta_i$ is the angle between $a_i$ and the space determined by $a_{i+1},\ldots,a_n$.

PS. Recall that the volume ${\rm vol}(a_i,\ldots,a_n)$ is defined by $${\rm vol}(a_i,\ldots,a_n)=\sqrt{\det ({A_i}^*A_i)},$$ where $A_i$ is the matrix with columns $a_i,\ldots,a_n$.

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John B
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