What is the maximum number of free independent real parameters specifying an
n-dimensional orthogonal matrix?
Options:
(A) n(n - 2)
(B) (n - 1)2
(C) n(n - 1)
2
(D) n(n + 1)
2
Official Answer Key:
C
Step-by-Step Solution & Explanation
### **Official Solution & Step-by-Step Derivation (GATE PH 2022 Q.13)**
**1. Group Theory & Mathematical Principles**:
- An $n$-dimensional real orthogonal matrix $O \in O(n)$ satisfies:
$O^T O = \mathbb{I}_n$
- Total entries in an $n \times n$ real matrix: $n^2$.
**2. Detailed Step-by-Step Derivation**:
- The matrix equation $O^T O = \mathbb{I}_n$ represents a system of scalar equations:
$\sum_{k=1}^n O_{ki} O_{kj} = \delta_{ij}, \quad 1 \le i, j \le n$
- Since $O^T O$ is inherently symmetric ($(O^T O)^T = O^T O$), the number of independent constraints is given by the number of independent elements in a symmetric $n \times n$ matrix:
$\text{Independent constraints} = \frac{n(n+1)}{2}$
- The number of free independent real parameters is the total parameters minus the number of independent constraints:
$\text{Free parameters} = n^2 - \frac{n(n+1)}{2} = \frac{2n^2 - n^2 - n}{2} = \frac{n(n - 1)}{2}$
- Physically, this corresponds to the dimension of the Lie group $SO(n)$, representing the $\frac{n(n-1)}{2}$ independent planes of rotation in $n$ dimensions.
**3. Official Verification**:
The verified official IIT Master Answer is **Option (C)** (`n(n - 1)/2`).
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