# What does inner product conjugate symmetry mean?

I know that the property $\langle u, v \rangle = \overline{\langle v, u \rangle}$ for inner product holds.

I have a matrix $A$, such that,

$\langle Ax, x \rangle = -\langle x, Ax \rangle$ i.e.

$\langle x, Ax \rangle = -\overline{\langle x, Ax \rangle}$.

What does this imply about $Ax$ and $x$?

I need to show that $\langle Ax, x \rangle = 0$, which is actually valid for any $A = -A^T$, which I verified using sample matrices, but cannot verify in general. Can anyone help?

Thanks.

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Let $E=\{e_i:i=\overline{1,n}\}$ be orthonormal basis of vector space $V$ over $\mathbb{C}$. Consider linear operator $A\in\mathcal{L}(V,V)$ such that $\langle Ax, x\rangle=-\langle x, Ax\rangle=-\overline{\langle Ax, x\rangle}$. Let $[a_{ij}]\in\mathbb{Mat}(\mathbb{C}^n)$ be the matrix of operator $A$ in basis $E$, then $a_{ij}=\langle Ae_i, e_j\rangle$, and $a_{ij}=-\overline{a_{ij}}$. For complex numbers the equality $z=-\overline{z}$ implies $z=0$, so $a_{ij}=0$ for all $i,j$. Since the matrix $[a_{ij}]=0$, then $A=0$.
From my proof you see that this is impossible: $\langle Ax,x\rangle=-\overline{\langle Ax,x\rangle}$ implies $A=0$. – Norbert Feb 1 '12 at 9:35