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

For a real n×n matrix M, the null space (kernel) is the set of vectors x with Mx = 0. Two governing facts follow directly from linear algebra: (1) M is singular — equivalently det(M) = 0 — exactly when its null space contains a nonzero vector, since a nonzero solution of Mx = 0 means the columns of M are linearly dependent; and (2) the null space is itself a linear subspace, so it is closed under scalar multiplication — once it holds one nonzero vector, it holds every nonzero scalar multiple of that vector.

1. The vector (0, 1, 0, …, 0) is the second standard basis vector, e2. It is nonzero and, by hypothesis, lies in the null space of M, so M e2 = 0.

2. Multiplying M by e2 extracts the second column of M, so M e2 = 0 means the second column of M is the zero column.

3. A matrix with a zero column has linearly dependent columns, so by fact (1) it is singular — hence det(M) = 0 for every such M.

4. By fact (2), since e2 is a nonzero vector in the null space, every nonzero scalar multiple c·e2 (c ≠ 0) is also in the null space — giving infinitely many nonzero null-space vectors, so certainly at least two.

Cross-check against the other claims:

• A determinant of 1 would require M to be invertible, but an invertible matrix cannot have a nonzero vector in its null space — so this value is impossible here, consistent with fact (1) instead forcing det(M) = 0.

• The rank is not pinned to any single number: only the second column is forced to be zero, while the remaining n − 1 columns can still be chosen linearly independent, so the rank can be as large as n − 1.

So the properties that hold for every matrix M satisfying the given condition are: det(M) = 0, and the null space contains at least two nonzero vectors.