GATE 2024 DA Q.47: solution and answer

Official answer: A, CSource: Official answer key, GATE 2024 DA (IISc Bengaluru)

1. A subset $S\subseteq {\mathbb{R}}^3$ is a subspace if it contains $0$ and is closed under addition and under scalar multiplication. Test each option.

2. (A) The set is $\mathrm{span}\left\{\left[\begin{array}{c}1\\ 1\\ 0\end{array}\right],\left[\begin{array}{c}1\\ 0\\ 0\end{array}\right]\right\}$, because $\alpha ,\beta$ range over all of $\mathbb{R}$. A span of vectors is always a subspace, so (A) is a subspace (it is the plane $x_3=0$).

3. (B) Here $x_1={\alpha }^2+{\beta }^2$, $x_2=2{\alpha }^2$, $x_3={\beta }^2$, so every vector in the set has $x_1\ge 0$, $x_2\ge 0$ and $x_3\ge 0$. Take $\mathbf{u}=\left[\begin{array}{c}1\\ 2\\ 0\end{array}\right]$ (with $\alpha =1,\beta =0$), which is in the set. Then $-\mathbf{u}=\left[\begin{array}{c}-1\\ -2\\ 0\end{array}\right]$ has $x_1<0$, so it is not in the set. Closure under scalar multiplication fails, so (B) is not a subspace. The squares allow only non-negative multipliers, giving a cone rather than a subspace.

4. (C) The set is the solution set of the homogeneous system $A\mathbf{x}=0$ with $A=\left[\begin{array}{ccc}5& 0& 2\\ 4& -2& 3\end{array}\right]$, that is, the null space of $A$. If $A\mathbf{x}=0$ and $A\mathbf{y}=0$, then $A(a\mathbf{x}+b\mathbf{y})=0$, so a null space is always a subspace. Hence (C) is a subspace (here a line: the two rows are independent, so the rank is 2 and the nullity is $3-2=1$).

5. (D) Put $\mathbf{x}=0$: $5(0)+2(0)+4=4\ne 0$. The zero vector is not in the set, so (D) is not a subspace. It is a plane that does not pass through the origin.

6. Hence the subspaces are (A) and (C).