91Ó°ÊÓ

Since the trace of a matrix is the sum of its diagonal elements, we need to find matrices in W with diagonal elements adding up to zero. One simple way to achieve this is to ensure that elements in the diagonal of the matrix are 0, leaving the other elements as arbitrary values. For example, we can consider the following matrices: Matrix A: \(\begin{bmatrix} 0 & 1 & 0 & \cdots & 0 \\ 0 & 0 & 0 & \cdots & 0 \\ 0 & 0 & 0 & \cdots & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & 0 & \cdots & 0 \end{bmatrix}\) Matrix B: \(\begin{bmatrix} 0 & 0 & 1 & \cdots & 0 \\ 0 & 0 & 0 & \cdots & 0 \\ 0 & 0 & 0 & \cdots & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & 0 & \cdots & 0 \end{bmatrix}\) and so on, until we reach Matrix \(M_{ij}\), where the only non-zero element is in the i-th row and j-th column, with i not equal to j.

The set of linearly independent matrices we found in Step 1 should form the basis of W if they span the entire subspace. To verify this, consider any arbitrary matrix in W. Let's call it Matrix X: Matrix X: \(\begin{bmatrix} 0 & a_{12} & a_{13} & \cdots & a_{1n} \\ a_{21} & 0 & a_{23} & \cdots & a_{2n} \\ a_{31} & a_{32} & 0 & \cdots & a_{3n} \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ a_{n1} & a_{n2} & \cdots & \cdots & 0 \end{bmatrix}\) We can notice that Matrix X can be written as a linear combination of the matrices we identified in Step 1.

To find the dimension of W, we need to count the number of matrices in the basis we found in Step 1. Since we have one matrix for every off-diagonal element of an \(n \times n\) matrix, there are \(n^2 - n\) such elements. Therefore, the dimension of W is \(n^2 - n\).