Discussion
Let \(e_1, e_2, e_3, e_4, e_5\) denote the standard basis vectors of the real vector space \(\mathbb{R}^{5}\). Let \(V\) be the subspace of \(\mathbb{R}^{5}\) spanned by the vectors \(e_1 + e_2, e_2 + e_3, e_3 + e_4 + e_5\). Consider the real vector space \(M_{2,5}(\mathbb{R})\). Let
\(S = \{P \in M_{2,5}(\mathbb{R}) \mid \text{nullspace}(P) = V\}\).
Which ONE of the following statements is TRUE?
Answer: C
The TRUE statement is: \(S\) is non-empty and \(S\) is not a subspace of \(M_{2,5}(\mathbb{R})\).
The three spanning vectors are linearly independent, so \(\dim V = 3\). A matrix \(P \in M_{2,5}(\mathbb{R})\) has rank at most 2, so by rank-nullity its null space has dimension \(5 - \text{rank}(P) \geq 3\). It equals \(V\) exactly when \(P\) has rank 2 and its two rows span the orthogonal complement \(V^{\perp}\), which has dimension 2.
Such matrices exist: the rows \((1,-1,1,-1,0)\) and \((1,-1,1,0,-1)\) are independent and orthogonal to all three spanning vectors. So \(S\) is not empty.
The zero matrix is not in \(S\), because its null space is all of \(\mathbb{R}^{5}\), not \(V\). A subspace must contain 0, so \(S\) is not a subspace, and \(S = \{0\}\) is wrong for the same reason. Closure fails too: if \(P \in S\) then \(-P \in S\), but \(P + (-P) = 0 \notin S\).
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