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

Step 1: Understand what it means for a vector to be in the null space.
The null space (kernel) of a matrix \( M \) is the set of all vectors \( v \) such that \( Mv = 0 \). We are told \( e_2 = (0,1,0,\ldots,0) \) satisfies \( M e_2 = 0 \), and since \( e_2 \) is not the zero vector, the null space of \( M \) contains at least one non-zero vector.

Step 2: Evaluate options (A) and (B) using the invertibility criterion.
A square matrix \( M \) is invertible (equivalently \( \det(M) \neq 0 \)) if and only if its null space contains only the zero vector. Since the null space of \( M \) contains the non-zero vector \( e_2 \), \( M \) cannot be invertible, so \( \det(M) = 0 \) always. This makes option (B) always TRUE, and rules out option (A) (determinant 1 is impossible here), so (A) is always FALSE.

Step 3: Evaluate option (C).
A non-trivial null space only guarantees nullity(M) is at least 1, i.e. rank(M) = n - nullity(M) is at most n-1; it does not fix the rank at exactly 1. For example, if \( M \) is the zero matrix, every vector including \( e_2 \) lies in its null space, yet rank(M) = 0, not 1. So (C) is not always correct.

Step 4: Evaluate option (D).
The null space of any matrix is a subspace of \( \mathbb{R}^n \), and a subspace is closed under scalar multiplication. Since \( e_2 \) is a non-zero vector in the null space, for any non-zero scalar \( c \), \( c \cdot e_2 \) is also in the null space and non-zero. Taking \( c=1 \) and \( c=2 \) gives two distinct non-zero vectors \( e_2 \) and \( 2e_2 \), both in the null space (in fact there are infinitely many). So (D) is always TRUE.

Step 5: Conclusion.
The statements that are always correct are (B) determinant of \( M \) is 0, and (D) there are at least two non-zero vectors in the null space of \( M \).

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