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Generic square matrix with positive 1 bounded entries

Considering a matrix $A=(a_{i.j})$ where $0 \leq a_{i,j} < 1 \forall i,j$. It is important to consider that all entries are strictly lower than 1 and positive.

Rows sum to a number lower than 1

Let us consider that the sum of all entries of matrix $A$'s rows is lower than 1: $\sum_{j=1}^{n}a_{i,j} < 1$. Sorry, maybe I did not specify it, only wrote in the formula, I talk about rows. Rows sum to a number lower than 1.


Let us consider $\det(A)$ (determinant).

Is it true that $\det(A)<1$???

Or maybe $|\det(A)| < 1$???

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Perhaps you want to put absolute value signs around everything? Otherwise, consider $\begin{bmatrix}-100 & 0 \\ 0 & -100\end{bmatrix}$. – Rahul Jun 4 '12 at 2:09
Yeah, all entries are positive. Please forgive me... – Andry Jun 4 '12 at 2:13
up vote 2 down vote accepted

EDIT: Taking into account the condition that the sum of the entries be less than one, the determinant is a sum of $n!$ terms, each of which is at most $n^{-n}$, so the determinant is bounded by $n!/n^n$, which is certainly less than 1 (for $n\gt1$). Each term is les than $n^{-n}$ because it's a product of $n$ numbers that add up to less than 1, and you maximize the product by taking all the number to equal $1/n$.

(Never mind --- I just saw the part about the sum of all the entries, or maybe it's the sum of all the entries in each row, being less than 1.)

Are we only talking about $2\times2$ matrices? If not, then $$\pmatrix{a&b&0\cr0&c&d\cr e&0&f\cr}$$ will have determinant $acf+bde$ which can certainly exceed 1 even if all the variables stand for numbers between zero and one.

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I am talking about sum of row entries... In your case, did you take into account this in your explaination? Sorry I am still reading, wanna be sure we are talking about the same. In my case I consider sum of row entries, not all elements in the matrix – Andry Jun 4 '12 at 2:52
I edited my questino because I specified the condition on rowas only in the frmula and not using words... It could be misleading. Very sorry for my carelessness – Andry Jun 4 '12 at 2:54
Why don't you go away and think about your question for a couple of days and come back when you're able to put into writing the actual question you want to ask instead of something with lots of conditions missing or incorrectly stated? While you're at it, look up the Hadamard bound on determinants, it might answer your question, depending, of course, on what the heck your question might really be. – Gerry Myerson Jun 4 '12 at 3:07
I can understand that due to my carelessness you had to go through some troubles in understanding what I was looking for. Actually the question is the one here now, no more edits. I simply had in my mind the matrix structure but failed in explaining it and providing good details. I apologized. This being said, I do not think to deserve your bad words as there are many other members in this community behaving really bad towards those who answer their questions. You could simply say to pay more attention, it would have been "more professional". – Andry Jun 4 '12 at 3:37
Each term in the sum isn't necessarily bounded by $n^{-n}$, take a constant multiple of the identity matrix for example.. making all terms ${1 \over n}$ actually minimizes things as the determinant becomes zero. – Zarrax Jun 4 '12 at 4:59

Note that $\sum_j a_{ij}^2 < \sum_j a_{ij} < 1$. So the magnitude of each row, viewed as a vector in ${\mathbb R}^n$, is less than one. The absolute value of the determinant of $A$ is the volume of the parallelopiped spanned by the rows, which is at most the product of the magnitudes of the row vectors, and therefore is less than one in this case.

If you want to do it algebraically, you can prove it by induction on the dimension, the $1$ by $1$ case being trivial. Then you can do a cofactor expansion along any $i$th row, getting that $$det(A) = \sum_j (-1)^{i + j} a_{ij} \,det(A_{ij})$$ Note that each matrix $A_{ij}$ also satisfies the conditions of the problems, so each $|det(A_{ij})| < 1$ by induction hypothesis. You then get $$|det(A)| < \sum_j |a_{ij}||det(A_{ij})|$$ $$< \sum_j |a_{ij}|$$ $$< 1$$

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I think you'll find that what you are using in the first paragraph is what's commonly referred to as the Hadamard bound. – Gerry Myerson Jun 4 '12 at 6:04

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