TITLE: What is a basis for the space of $n\times n$ Hermitian matrices?
QUESTION [7 upvotes]: So I was working on a specific problem related to Hermitian matrices. If we let $H_n$ denote the set of n x n Hermitian matrices. We're told that $H_n$ is a real vector space under matrix addition and scalar multiplication by a real number. I don't understand why though, because what would prevent you from adding two imaginary numbers? 
Then we're told to write a basis for $H_2$. Now for a Hermitian matrix, it's equal to its conjugate transpose, so a basis would be:
$$\left(\begin{array}{ccc}
0 & 1 \\
1 & 0\\
\end{array}\right) , \left(\begin{array}{ccc}
0 & 0 \\
0 & 1\\
\end{array}\right), \left(\begin{array}{ccc}
1 & 0 \\
0 & 0\\
\end{array}\right), \left(\begin{array}{ccc}
0 & i \\
-i & 0\\
\end{array}\right)$$
Which should be the same as the basis for a symmetric matrix, correct? 
From there though, we have $sH_n$ represent skew-Hermitian matrices. I believe the basis for $sH_n$ is:
Edit: updated my basis here. 
$$\left(\begin{array}{ccc}
0 & -1 \\
1 & 0\\
\end{array}\right), \left(\begin{array}{ccc}
0 & i \\
i & 0\\
\end{array}\right)$$
because the diagonals would have to be zero right, to make the conjugate transpose of A be equal to -A?
From there though, they ask if $sH_n$ is a real vector space, and I'm not sure what to answer. It's real if we only use reals?
I'm asked the same thing for $U_n$, where $U_n$ represents n x n unitary matrices. I'm not even sure how to come up with a basis for this set??
Edit: Technically, the question says "Is $U_n$ a real vector space? Write a basis if possible."
I suppose I couldn't write a basis because $U_n$ is not a real vector space then. How would I show that though? 
Thanks in advance for your help.

REPLY [6 votes]: In the general case, a standard basis is given (up to normalisation) by
$$E_{ii},\qquad E_{ij}+E_{ji}, \qquad i(E_{ij}-E_{ji}),$$
for all $i,j=1,...,n$ with $i<j$, and where I used the notation $E_{ij}\equiv e_i e_j^*$, where $e_i$ is the unit vector that is nonzero only in its $i$-th component: $(e_i)_j =\delta_{ij}$ and $(E_{ij})_{k\ell}=\delta_{ik}\delta_{j\ell}$.
Clearly all these matrices are Hermitian. To verify that form an orthogonal basis, we just need to check that
$$\mathrm{Tr}(E_{ii} (E_{jk}+E_{kj}))
=\mathrm{Tr}(E_{ii}(E_{jk}-E_{kj}))) 
=\mathrm{Tr}(E_{ii}E_{jk})) = \delta_{ij}\delta_{ik} = 0, \qquad (\text{as } j<k), \\
\mathrm{Tr}[(E_{ij}+E_{ji})(E_{k\ell}-E_{\ell k})]
= 0, \\
\mathrm{Tr}[(E_{ij}+E_{ji})(E_{k\ell}+E_{\ell k})]
=2(\delta_{jk}\delta_{i\ell} + \delta_{j\ell}\delta_{ik}), \\
\mathrm{Tr}[(E_{ij}-E_{ji})(E_{k\ell}-E_{\ell k})]
=2(\delta_{jk}\delta_{i\ell} - \delta_{j\ell}\delta_{ik}).$$

See also this related question.