Question

Let \( n > 1 \). Consider an \( n \times n \) matrix \( M \) with its elements from \( \mathbb{R} \). Let the vector \( (0, 1, 0, 0, \ldots, 0) \in \mathbb{R}^n \) 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 non-zero vector in the null space makes the matrix singular; also remember a null space is a subspace, so it is closed under scalar multiplication.

Answer 1

The Correct Option is B, D

Method: Test every option using the Rank-Nullity Theorem and a concrete counterexample matrix.
The Rank-Nullity theorem states rank(M) + nullity(M) = n. We are given nullity(M) is at least 1 since \( e_2 \neq 0 \) lies in the null space.

Testing (A): determinant = 1 would mean M is invertible, i.e. nullity(M) = 0, contradicting nullity(M) at least 1. So (A) can never hold.

Testing (B): a square matrix is non-invertible exactly when its determinant is 0. Since nullity(M) is at least 1 always, det(M) = 0 always. (B) holds unconditionally.

Testing (C) with a counterexample for \( n=3 \): let \( M = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix} \). Here \( M(0,1,0)^T = (0,0,0)^T \), so the condition holds, and rank(M) = 1 here. But for the zero matrix \( M = 0 \), the same vector is also in the null space, yet rank(M) = 0, not 1. Since the option must hold for every matrix satisfying the condition and it fails for the zero matrix, (C) is not always correct.

Testing (D): the null space is a subspace, closed under scalar multiplication, so it always contains \( e_2, 2e_2, 3e_2, \ldots \), i.e. infinitely many non-zero vectors, certainly at least two. (D) holds unconditionally.

So exactly (B) and (D) hold in every case.

\[ \boxed{\text{Options (B) and (D)}} \]