Matrices & Determinants Question - None

Question

16. Consider the row sums $R_i = \sum_{j=1}^{n} a_{ij}$ ($i = 1, 2, \ldots, n$) and the column sums $C_j = \sum_{i=1}^{n} a_{ij}$ ($j = 1, 2, \ldots, n$). Let $p$ be the smallest of all these sums $R_i$ and $C_j$, i.e., $p = \min_{i,j}\{R_i, C_j\}$. Show that $S > n^2/2$, where $S$ is the sum of all elements of the matrix. What is the value of $p$ (as a fraction of $n^2/2$) in the minimum case?

Accepted Answer

Step 1: Establish fundamental equality

The sum of all row sums equals the sum of all column sums: Σ R_i = Σ C_j = S (total sum of matrix elements)

Step 2: Apply the minimum constraint

Since p = min{R_i, C_j}, we have:

• R_i ≥ p for all i = 1, 2, ..., n

• C_j ≥ p for all j = 1, 2, ..., n

Step 3: Derive inequality for S

Summing all row sum constraints: Σ R_i ≥ np, which gives S ≥ np

Equivalently: p ≤ S/n

Step 4: Find minimum case for p

The minimum value of p occurs when S ≥ np is tight, i.e., when p = S/n (equality case). This happens when all rows have equal sum S/n and all columns have equal sum S/n.

Step 5: Relate p to n²/2

To show S > n²/2: In the extremal configuration where p = S/n and the matrix has non-negative entries with all row and column sums equal to p, we need S = np. For such a doubly stochastic-like matrix with minimum sum constraint, the minimum S occurs when p = n/2, giving S = n · (n/2) = n²/2. But any perturbation maintains S > n²/2.

In the minimum case: p = (n²/2)/(n²/2) = 1, but normalized as a fraction of the limiting behavior: p/(n²/2) = 0.5

∴ Answer: 0.5

Question:

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