If a matrix can be written as A = uvT, where both u and v are n-dimensional real-
valued non-zero column vectors, then the rank of the matrix A is ________
(answer in integer).
Official Answer Key:
1.0
Step-by-Step Solution & Explanation
### **Detailed Step-by-Step Solution (GATE AE 2026 Q.32)**
**1. Intuitive Physical Setup & Given Parameters**:
Let $A = u v^T$, where:
- $u$ is an $n$-dimensional real non-zero column vector ($n imes 1$).
- $v$ is an $n$-dimensional real non-zero column vector ($n imes 1$).
- $A$ is an $n imes n$ matrix (the outer product of $u$ and $v$).
We need to determine the rank of matrix $A$.
**2. Step-by-Step Mathematical Derivation**:
1. **Structure of the Outer Product Matrix**:
- Let $u = egin{bmatrix} u_1 \ u_2 \ dots \ u_n \end{bmatrix}$ and $v = egin{bmatrix} v_1 \ v_2 \ dots \ v_n \end{bmatrix}$.
- The matrix $A = u v^T$ is given by:
A = egin{bmatrix} u_1 v_1 & u_1 v_2 & \cdots & u_1 v_n \ u_2 v_1 & u_2 v_2 & \cdots & u_2 v_n \ dots & dots & \ddots & dots \ u_n v_1 & u_n v_2 & \cdots & u_n v_n \end{bmatrix} = egin{bmatrix} v_1 u & v_2 u & \cdots & v_n u \end{bmatrix}
2. **Column Space & Rank**:
- Every column of $A$ is a scalar multiple of the vector $u$:
\text{Column } j \text{ of } A = v_j u
- Since $v
eq 0$, at least one $v_j
eq 0$, meaning at least one column is non-zero.
- All columns are linearly dependent and span a 1-dimensional subspace:
\operatorname{Col}(A) = \operatorname{span}\{u\}
- Therefore, the dimension of the column space (the rank of $A$) is exactly 1:
\operatorname{rank}(A) = 1
**3. Comprehensive Options & Distractor Analysis**:
- **Evaluation Criteria**: Numerical Answer Type (NAT).
- **Exact Analytical Result**: $\operatorname{rank}(A) = 1$.
- **Official Tolerance Range**: Integer answer: **1**.
**4. Official Master Key & Exam Takeaway**:
- **Official Answer Key**: **1.0** (Acceptable Range: **1 to 1**)
- **Exam Takeaway**: Any non-zero outer product matrix $u v^T$ has rank 1. In fact, by Frobenius inequality, $\operatorname{rank}(u v^T) \le \min(\operatorname{rank}(u), \operatorname{rank}(v^T)) = 1$.