Answer to Question #155700 in Linear Algebra for Roshan

Question

Let V = Mn(C) and U be the subspace of all Hermitian matrices over C. Find a basis and

dimension of U.

Answer 1

Hermitian matrix is a matrix A, which has components a_ij, that satisfy the condition $a_{ij} = \bar{a}_{ji}$. This condition is invariant under multiplication by real numbers and not invariant under multiplication by complex numbers.

Therefore, U is not a complex subspace of M_n(C), but a real subspace only.

The real basis of U may be composed of the matrices of three types:

1) D_k, (k =1, 2, ..., n) which has zeros in all the cells besides the intersection of k-th column and k-th row, where it has 1. There are n matrices of this type.

2) E_kl, (k,l = 1, 2, ..., n, k>l) which has zeros in all the cells besides the intersection of k-th column and l-th row and the intersection of l-th column and k-th row, where it has 1. There are n(n-1)/2 matrices of this type.

3) I_kl, (k,l = 1, 2, ..., n, k>l) which has zeros in all the cells besides the intersection of k-th column, where it has i, and l-th row and the intersection of l-th column and k-th row, where it has -i. There are n(n-1)/2 matrices of this type.

The total amount of the matrices in this basis is n + n(n-1)/2 +n(n-1)/2 = n².

Answer. $\dim_{\mathbb{R}} U = n^2$