Question

Let W be the subspace of \(M_3(\R)\) consisting of all matrices with the property that the sum of the entries in each row is zero and the sum of the entries in each column is zero. Then the dimension of W is equal to ___________.

Answer 1

Correct Answer: 4

To determine the dimension of the subspace \( W \) of \( M_3(\mathbb{R}) \), consisting of all \( 3 \times 3 \) matrices where the sum of the entries in each row and each column is zero, we follow these steps:

1. Consider a \( 3 \times 3 \) matrix \( A = [a_{ij}]_{3 \times 3} \). For \( A \) to be in \( W \), each row sum must be zero, i.e., \( a_{i1} + a_{i2} + a_{i3} = 0 \) for all \( i \), and each column sum must be zero, i.e., \( a_{1j} + a_{2j} + a_{3j} = 0 \) for all \( j \).
2. Initially, a general \( 3 \times 3 \) matrix can have any real number for each entry, thus having \( 9 \) degrees of freedom.
3. Each of the 3 row sum conditions imposes 1 constraint: \( a_{i3} = -a_{i1} - a_{i2} \) for each \( i \). Similarly, each of the 3 column sum conditions imposes \( 1 \) more independent constraint since the last entry can be expressed in terms of the others.
4. Let's count the independent constraints: we have 3 independent constraints from row sums and 2 additional (because the third column constraint becomes dependent on the others) from the column sums. Total: 3 + 2 = 5 constraints.
5. Thus, the number of free variables is \( 9 - 5 = 4 \), which means the dimension of \( W \) is 4.

This result falls within the given range of 4 to 4. Thus, the dimension of \( W \) is confirmed to be 4.