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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 $\frac{n(n+1)}{2}$

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  • $\begingroup$ im not getting @JohnHughes,,can u elaborate more $\endgroup$
    – jasmine
    Jul 4 '18 at 15:40
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    $\begingroup$ See wikipedia under "properties". $\endgroup$ Jul 4 '18 at 15:47
  • $\begingroup$ @JohnHughes Seems correct for $n=1$ to me, if not in general... $\endgroup$
    – gj255
    Jul 4 '18 at 15:55
  • $\begingroup$ 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. $\endgroup$ Jul 5 '18 at 10:11
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I believe the best way to approach such a problem is to try come up with a basis for the vector space.

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 $n+\sum_{k=1}^{n-1}2k=n+2\frac{n(n-1)}{2}=n^2$

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