Question

Let 𝑛 > 1. Consider an 𝑛× 𝑛 matrix 𝑀 with its elements from ℝ. Let the vector
(0, 1, 0, 0, … , 0) ∈ℝ𝑛 be in the null space of 𝑀.
Which of the following options is/are always correct?

• A. Determinant of 𝑀 is 1
• B. Determinant of 𝑀 is 0
• C. Rank of 𝑀 is 1
• D. There are at least two non-zero vectors in the null space of 𝑀

Hint:

A nonzero vector in the null space forces the matrix to be singular (det = 0), and since null spaces are closed under scalar multiplication, any nonzero vector in it generates infinitely many nonzero vectors there too.

Answer 1

The Correct Option is B, D

Step 1: Interpret the given information. The vector \(v=(0,1,0,0,\ldots,0)\) is nonzero and lies in the null space of \(M\), i.e. \(Mv = 0\).

Step 2: Determine the determinant. A square matrix \(M\) has a nonzero null space if and only if \(M\) is singular, which means \(\det(M) = 0\). So \(\det(M)=0\) always holds - Option B is correct, and Option A (\(\det(M)=1\)) is false.

Step 3: Check the rank claim. By the rank-nullity theorem, \(\text{rank}(M) + \dim(\text{Null}(M)) = n\). We only know \(\dim(\text{Null}(M)) \ge 1\), so \(\text{rank}(M) \le n-1\). The rank could be \(n-1\), \(n-2\), or even \(0\) (e.g. the zero matrix), depending on \(M\). Nothing forces the rank to be exactly \(1\), so Option C is not always correct.

Step 4: Check whether the null space must contain at least two nonzero vectors. The null space of any matrix is a subspace of \(\mathbb{R}^n\). Since \(v \ne 0\) belongs to this subspace, every scalar multiple \(cv\) for \(c \in \mathbb{R}\) also belongs to it (subspaces are closed under scalar multiplication). Choosing two distinct nonzero scalars \(c_1 \ne c_2\) gives two distinct nonzero vectors \(c_1v\) and \(c_2v\) in the null space. So the null space always contains at least two nonzero vectors - Option D is correct.

\[\boxed{\text{Always correct: Options B and D}}\]