GATE Civil 2023 Set 2 | Question: 27

Question

Two vectors $\left[\begin{array}{llll}2 & 1 & 0 & 3\end{array}\right]^{T} $ and $\left[\begin{array}{cccc}1 & 0 & 1 & 2\end{array}\right]^{T}$ belong to the null space of a $4 \times 4$ matrix of rank $ 2.$ Which one of the following vectors also belongs to the null space? 

• A. $\left[\begin{array}{llll}1 & 1 & -1 & 1\end{array}\right]^{T}$
• B. $\left[\begin{array}{llll}2 & 0 & 1 & 2\end{array}\right]^{T}$
• C. $\left[\begin{array}{llll}0 & -2 & 1 & -1\end{array}\right]^{T}$
• D. $\left[\begin{array}{llll}3 & 1 & 1 & 2\end{array}\right]^{T}$

Answer 1

Step 1: Identify Null Space Vectors
Given vectors $u = [2, 1, 0, 3]^T$ and $v = [1, 0, 1, 2]^T$ belong to the null space.

Step 2: Apply Subspace Property:
Any linear combination $w = c_1u + c_2v$ must also belong to the null space.

Step 3: Test Linear Combinations
Subtract $v$ from $u$:
$u - v = [2-1, 1-0, 0-1, 3-2]^T = [1, 1, -1, 1]^T$ \\
This matches Option A.

Important Rule: The null space of a matrix is a vector subspace; it is closed under addition and scalar multiplication.

Comments

How do you know those vectors form a basis of the null space? Your entire answer rests on that conclusion only.

That needs to be proven using rank nullity. Since rank = 2, dimension of the null space is 2. And those two vectors are themselves linearly independent. Hence they form a basis of the null space.