What is a basis for the space of $n\times n$ Hermitian matrices? 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. 
 A: A basis for $H_2$ over $\mathbb{R}$ is
$$
\left(
\begin{array}{cc}
1&0\\
0&0\\
\end{array}
\right),
\left(
\begin{array}{cc}
0&0\\
0&1\\
\end{array}
\right),
\left(
\begin{array}{cc}
0&i\\
-i&0\\
\end{array}
\right),
\left(
\begin{array}{cc}
0&1\\
1&0\\
\end{array}
\right).
$$
Being a real vector space doesn't mean that the entries have to be real, just that it is closed under multiplication by real scalars.
Another way of phrasing this is that an arbitrary $2\times 2$ Hermitian matrix is of the form
$$
\left(
\begin{array}{cc}
a&c+di\\
c-di&b\\
\end{array}
\right) \text{ with } \ a, \ b, \ c, \ d\in\mathbb{R}.
$$
As noted in another answer, Hermitian matrices do not form a complex vector space, i.e. for $\alpha\in\mathbb{C}\setminus\mathbb{R}$ and Hermitian $H$
$$
\overline{\alpha H}=\bar{\alpha}\overline{H}=\bar{\alpha}H\neq\alpha H.
$$
A: 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.
A: *

*I take it your specific question in the first instance is: I don't understand why it has to be a real vector space because we can add together complex numbers just as well, can't we? Yes, but don't forget the little detail of complex conjugates. Observe that $$i\begin{pmatrix}0&i\\-i&0\end{pmatrix}+\begin{pmatrix}0&1\\1&0\end{pmatrix}=\begin{pmatrix}0&0\\2&0\end{pmatrix}$$ is a $\mathbb{C}$-linear combination of Hermitian matrices, and the result is not Hermitian.

*With entries strictly in $\mathbb{R}$, Hermitian matrices are just symmetric matrices so your basis is correct and is indeed the very one for symmetric matrices. However, the problem probably wants the matrices here to include components from $\mathbb{C}$, so what you have is incomplete. By linearity, try treating the real and imaginary parts of the elements of $H_n$ separately.

*As for the basis for what you call $SH_2$, what about the following matrix, which isn't represented by what you have? And do you know what "diagonals" refers to in a matrix? $$\begin{pmatrix}0&1\\-1&0\end{pmatrix}$$ Of course, this one probably wants complex hermitian matrices as well.

*Careful. The addition of two unitary matrices is generally not unitary, so try to understand what the question means...
