Question

Let 𝑛 > 1. Consider an 𝑛× 𝑛 matrix 𝑀 with its elements from ℝ. Let the vector
(0, 1, 0, 0, … , 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 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

Approach this using the core fact that a matrix null space is a vector subspace, together with the rank-nullity relationship.

Since the specific vector \(v = (0,1,0,\ldots,0)\) satisfies \(Mv=0\) and \(v\) is not the zero vector, \(M\) maps a nonzero vector to zero. Any square matrix that does this is, by definition, not invertible, and a non-invertible square matrix always has determinant exactly \(0\), never \(1\). This immediately kills option A and confirms option B.

For the rank: the rank-nullity theorem says \(\text{rank}(M) = n - \dim(\text{Null}(M))\). All we are told is that the nullity is at least \(1\); it could be exactly \(1\) (giving rank \(n-1\)), or much larger (giving a much smaller rank). Since the exact nullity is unspecified, we cannot pin the rank down to \(1\) in general, so option C need not hold.

For the null space size: null spaces are always closed under scalar multiplication because they are vector subspaces. So if \(v \ne 0\) is in the null space, then so is \(2v\), \(3v\), \(-v\), and in fact infinitely many scalar multiples of \(v\), all nonzero (as long as the scalar is nonzero) and all distinct from each other. This guarantees at least two nonzero vectors in the null space, confirming option D regardless of what \(M\) actually is.

So the statements that are always true are option B (determinant is \(0\)) and option D (at least two nonzero vectors in the null space).