Linear Algebra Done Right · Problem 7A.18

To show a set isn't a subspace, you need to find a concrete failure of one of the subspace axioms—usually closure under addition or scalar multiplication. The key insight is that normality ($T^*T = TT^*$) is a multiplicative constraint, not a linear one, so two normal operators can easily fail to commute with each other's adjoints when added together. Try constructing two simple normal operators (like commuting diagonalizable ones) whose sum is not normal by checking whether $(T_1 + T_2)^*(T_1 + T_2) = (T_1 + T_2)(T_1 + T_2)^*$.

To show a set isn't a subspace, you need to find a concrete failure of one of the subspace axioms—usually closure under addition or scalar multiplication. The key insight is that normality ($T^*T = TT^*$) is a multiplicative constraint, not a linear one, so two normal operators can easily fail to commute with each other's adjoints when added together. Try constructing two simple normal operators (like commuting diagonalizable ones) whose sum is not normal by checking whether $(T_1 + T_2)^*(T_1 + T_2) = (T_1 + T_2)(T_1 + T_2)^*$.