3.1.1

Consider the vector space $$V=M_{nxn}(\mathbb{R})$$, consisting of real $$nxn $$ matrices, over the field $$\mathbb{R} $$.

a) The set of symmetric matrices forms a subspace of V. Let A and B be symmetric matrices, and let $$A_{ij} $$ denote the entry in row i and column j of A, then for i and j satisfying $$1 \le i,j \le n $$, we have $$A_{ij} = A_{ji} $$ and similarly for B. Then, $$(A+B)_{ij} = A_{ij} + B_{ij} = A_{ji} + B_{ji} = (A+B)_{ji} $$, so A+B is also symmetric, and this set is closed under addition. Now let c be a real number, then $$(cA)_{ij} = c(A_{ij}) = c(A_{ji}) = (cA)_{ji} $$, so cA is also symmetric, thus this set is closed under scalar multiplication and hence forms a subspace.

b) The set of invertible matrices does not form a subspace of V. Let A be any invertible matrix, then multiplication of A by 0(the zero scalar) gives the zero matrix, which is not invertible, thus this set is not closed under scalar multiplication and so does not form a subspace.

c) The set of upper triangular matrices forms a subspace of V. Let A and B be upper triangular matrices, then for i and j satisfying $$1 \le j < i \le n $$, $$A_{ij} = B_{ij} = 0 $$. That is, the entries of A and B below the diagonal are 0. Since the entries of A+B are obtained by adding the corresponding entries of A and B, they are also 0 below the diagonal, thus A+B is also upper triangular and the set is closed under addition. Similarly, multiplication of A by a scalar c to form cA will also result in cA having entries of 0 below the diagonal, thus this set is closed under scalar multiplication and hence forms a subspace.