Question

Let \( V \) be a nonzero subspace of the complex vector space \( M_7(\mathbb{C}) \) such that every nonzero matrix in \( V \) is invertible. Then, the dimension of \( V \) over \( \mathbb{C} \) is:

• A. 1
• B. 2
• C. 7
• D. 49

Hint:

A subspace of invertible matrices must be spanned by one invertible matrix, as invertibility is a property of a single matrix, leading to a one-dimensional subspace.

Answer 1

The Correct Option is A

The space \( M_7(\mathbb{C}) \) is the set of all \( 7 \times 7 \) matrices with complex entries. The dimension of this space over \( \mathbb{C} \) is \( 49 \) because each element of the matrix has 7 rows and 7 columns, thus \( 7 \times 7 = 49 \) entries. The subspace \( V \) is a subspace of \( M_7(\mathbb{C}) \), and every nonzero matrix in \( V \) is invertible. This implies that the nonzero matrices in \( V \) must form a space of invertible matrices. The set of invertible matrices in \( M_7(\mathbb{C}) \) is an open subset of \( M_7(\mathbb{C}) \) with respect to the topology induced by the norm, and this open set can only be spanned by a single matrix when the subspace is finite-dimensional. Thus, the dimension of \( V \) over \( \mathbb{C} \) must be 1, corresponding to the fact that \( V \) is spanned by a single invertible matrix. Therefore, the correct answer is (A).