Question

Let V be a nonzero subspace of the complex vector space 𝑀₇(ℂ) such that every nonzero matrix in 𝑉 is invertible. Then, the dimension of V over ℂ is

• A. 1
• B. 2
• C. 7
• D. 49

Answer 1

The Correct Option is A

To find the dimension of the subspace \(V\) of the complex vector space \(M_7(\mathbb{C})\), where every nonzero matrix in \(V\) is invertible, we will reason through the properties of matrices and subspaces.

1. **Understanding the Vector Space**: 

• \(M_7(\mathbb{C})\) denotes the space of all \(7 \times 7\) matrices where each entry is a complex number.
• The dimension of \(M_7(\mathbb{C})\) is \(7 \times 7 = 49\).

2. **Condition on Subspace \(V\)**:

• Every nonzero matrix in \(V\) is invertible.
• An invertible matrix has a determinant that is non-zero.
• In terms of linear algebra, a subspace where every element is invertible suggests that all nonzero elements also form a linearly independent set.

3. **Inferring the Dimension**:

• Suppose the dimension is greater than 1, then we could form linear combinations of matrices in \(V\) to produce a zero matrix, which cannot be invertible, violating the subspace condition since a linear combination that yields zero signals dependent vectors.
• Therefore, to satisfy the condition that every nonzero element of \(V\) is invertible, \(V\) must be spanned by only one invertible matrix. Hence, the dimension of \(V\) must be 1.

4. **Conclusion**:

• Thus, the dimension of \(V\) over \(\mathbb{C}\) is 1.

Therefore, the correct answer is 1.