# Symmetric, skew-symmetric $\mathbb{R}$-bilinear form, matrix of form in $\mathbb{R}$-basis.

Let $(-, -): V \times V \to \mathbb{C}$ be a positive definite Hermitian inner product on a finite dimensional $\mathbb{C}$-vector space $V$ and let $e_1, \ldots, e_n$ be an orthonormal basis of $V$. For any $v$, $w \in V$, let $(v, w)_{\text{Re}}$ be the real, resp. $(v, w)_{\text{Im}}$ the imaginary, part of the complex number $(v, w)$. Thus, we have$$(v, w) = (v, w)_{\text{Re}} + i \cdot (v, w)_{\text{Im}}.$$Let $V_\mathbb{R}$ be $V$ viewed as a vector space over $\mathbb{R}$, so $e_1, \ldots, e_n, i \cdot e_1, \ldots, i \cdot e_n$, is an $\mathbb{R}$-basis of $V_\mathbb{R}$.

1. How do I see that $v$, $w \mapsto (v, w)_\text{Re}$, resp. $v$, $w \mapsto (v, w)_{\text{Im}}$, is a symmetric, resp. skew-symmetric, $\mathbb{R}$-bilinear form on $V_\mathbb{R}$?
2. What is the matrix of that bilinear form in the above $\mathbb{R}$-basis?
• Hint: $\Re(z) = (z + \bar z) / 2$ – user251257 Sep 19 '16 at 18:42

How do I see that $v$, $w \mapsto (v, w)_\text{Re}$, resp. $v$, $w \mapsto (v, w)_{\text{Im}}$, is a symmetric, resp. skew-symmetric, $\mathbb{R}$-bilinear form on $V_\mathbb{R}$?
Since $$(v,w)=\overline{(w,v)}=\overline{(w,v)_\mathrm{Re}+i(w,v)_\mathrm{Im}}=(w,v)_\mathrm{Re}-i(w,v)_\mathrm{Im},$$ we have $$(v,w)_\mathrm{Re}+i(v,w)_\mathrm{Im}=(w,v)_\mathrm{Re}-i(w,v)_\mathrm{Im}$$ that is $$(v,w)_\mathrm{Re}=(w,v)_\mathrm{Re},\quad (v,w)_\mathrm{Im}=-(w,v)_\mathrm{Im}.$$
What is the matrix of that bilinear form in the above $\mathbb{R}$-basis?
Find it directly. What is in the $(j,k)$ cell of the matrix?
• @JakobW: Denote $(u_1,\ldots,u_{2n})=(e_1,\ldots,e_n,ie_1,\ldots ie_n)$, then the matrix of $(-,-)_\mathrm{Re}$ in the $(j,k)$-cell contain $(u_j,u_k)_\mathrm{Re}$. It can be easy found because $e_1,\ldots,e_n$ is an orthonormal basis. Similarly for $(-,-)_\mathrm{Im}$. – Canis Lupus Sep 28 '16 at 15:54