# What is dimension over $\mathbb R$ of the space of $n\times n$ Hermitian matrices? [duplicate]

what is dimension over $$\mathbb{R}$$ of $$H_n( \mathbb{C})$$, the set of $$n \times n$$ Hermitian matrices?

My attempt: every real number is a complex number as all symmetric matrices are Hermitian. In my view the dimension of $$H_n( \mathbb{C})$$ is $$\frac{n(n+1)}{2}$$

• im not getting @JohnHughes,,can u elaborate more Commented Jul 4, 2018 at 15:40
• See wikipedia under "properties". Commented Jul 4, 2018 at 15:47
• @JohnHughes Seems correct for $n=1$ to me, if not in general... Commented Jul 4, 2018 at 15:55
• My comment should have been for $n = 2$, alas. So much for being smug. Still, I wish OP had tested out the conjecture for at least a couple of cases. Commented Jul 5, 2018 at 10:11

The diagonal elements of such a matrix must be real in order to be equal to their complex conjugate. So the diagonal Hermitian matrices are spanned by the $n$ vectors $\left(\begin{array}{ccc}1&&\\&\ldots&\\ &&0\end{array}\right),\left(\begin{array}{ccc}0&&\\&\ldots&\\ &&1\end{array}\right)$.
Every other entry below the diagonal is the complex conjugate of the corresponding element above the diagonal, so the matrix is determined by the rest of the elements above the diagonal. Over $\mathbb R$, the basis for $\mathbb C$ is simply the vectors $1$ and $i$. So there are two basis matrices for every position above the diagonal: the matrix which contains a $1$ at position $(j,k)$ and the matrix which contains an $i$ at position $(j,k)$.
This yields $n+\sum_{k=1}^{n-1}2k=n+2\frac{n(n-1)}{2}=n^2$