Solution: Minimum Elements for Skew-Symmetric Matrix

The question asks for the minimum number of elements required to uniquely define a 4x4 matrix ($n=4$) given two conditions:

1. All diagonal elements are zero ($m_{ii} = 0$).
2. The matrix is skew-symmetric ($m_{ij} = -m_{ji}$).

Matrix Properties Analysis

• Matrix Size: The matrix $M$ is $4 \times 4$, meaning it has $n^2 = 4^2 = 16$ total elements.
• Diagonal Elements ($m_{ii}$): The condition $m_{ii} = 0$ fixes all 4 diagonal elements ($m_{11}, m_{22}, m_{33}, m_{44}$) to zero. These do not need to be specified independently.
• Off-Diagonal Elements ($m_{ij}, m_{ji}$ where $i \neq j$): There are $n^2 - n = 16 - 4 = 12$ off-diagonal elements. The skew-symmetric condition $m_{ij} = -m_{ji}$ links these elements in pairs. For instance, $m_{12}$ and $m_{21}$ are related by $m_{12} = -m_{21}$.

Calculating Required Elements

To fully specify the matrix, we only need to determine the values that are not fixed by the conditions.

1. Diagonal elements: 0 elements need specification as they are all 0.
2. Off-diagonal elements: There are 12 off-diagonal elements, forming $\frac{n^2 - n}{2}$ pairs. For $n=4$, there are $\frac{16 - 4}{2} = \frac{12}{2} = 6$ pairs.
3. Since knowing one element in a skew-symmetric pair ($m_{ij}$) determines the other ($m_{ji} = -m_{ij}$), we only need to specify one element from each of the 6 pairs.
4. Therefore, the minimum number of elements required is 6.

General Formula Application

The number of independent elements required for an $n \times n$ skew-symmetric matrix is given by the formula $\frac{n(n-1)}{2}$. Plugging in $n=4$: $ \frac{4(4-1)}{2} = \frac{4 \times 3}{2} = \frac{12}{2} = 6 $