Question-15

Let \(\mathbb{R}^{n \times n}\) be the vector space of \(n \times n\) matrices with the usual matrix addition and scalar multiplication defined on matrices. Let \(\mathcal{L}\) be the subspace of all \(n \times n\) lower triangular matrices and \(\mathcal{U}\) be the subspace of all \(n \times n\) upper triangular matrices. Select all true statements.

• \(\mathcal{L} \cap \mathcal{U}\) contains just the zero matrix.
• The dimension of \(\mathcal{L}\) is \(\cfrac{n(n + 1)}{2}\)
• The dimension of \(\mathcal{U}\) is \(n^2\)
• \(\mathcal{L}\) and \(\mathcal{U}\) are isomorhpic subspaces of \(\mathbb{R}^{n \times n}\)

Answer

• \(\mathcal{L} \cap \mathcal{U}\) contains just the zero matrix.
• The dimension of \(\mathcal{L}\) is \(\cfrac{n(n + 1)}{2}\)
• The dimension of \(\mathcal{U}\) is \(n^2\)
• \(\mathcal{L}\) and \(\mathcal{U}\) are isomorhpic subspaces of \(\mathbb{R}^{n \times n}\)

Solution

In a lower(upper) triangular matrix, \(\cfrac{n(n + 1)}{2}\) elements can be non-zero and the rest have to be zero. The dimension of the space follows. Every upper triangular matrix in \(\mathcal{U}\) can be mapped to its transpose in \(\mathcal{L}\). This specifies an isomorphism between the two spaces. Every diagonal matrix is both upper triangular and lower triangular. Hence \(\mathcal{L} \cap \mathcal{U}\) is actually more numerous than just the singleton set. In fact, the intersection of these two subspaces is the subspace made up of all \(n \times n\) diagonal matrices.