# A conjugate LT in an inner product space

Let $V$ be a finite dimensional complex inner product space. Let $J$ be a conjugate linear map from $V$ to $V$ such that $J^2 =1$.

Can we say $\langle Ju, Jv \rangle = \langle v, u \rangle$ for all $u, v$ in $V$?

Conjugate linear means $J(u+v) = J(u) + J(v)$ and $J(a.v) = \overline{a}.J(v)$

• What is conjugate linear ? – HK Lee Mar 1 '16 at 12:27
• sorry J not linear operator may imply J(u+w)=J(u)+J(w) @user1952009. See your example itself there is no problem. – Sushil Mar 1 '16 at 12:56
• No I mean same. V is COMPLEX IPS. J is antilinear map(if you familiar with this word). and $J^2=I$. Can we say $<Ju, Jv> = <v , u>$ – Sushil Mar 1 '16 at 13:04
• en.wikipedia.org/wiki/Antilinear_map edit your question to include that and say that $J$ is NOT a linear operator $V \to V$ – reuns Mar 1 '16 at 13:07
• you know that $\mathbb{C}^n$ is naturally isomorphic to $\mathbb{R}^{2n}$ ? – reuns Mar 1 '16 at 13:17

It is in general wrong. Consider on $V=\mathbb C^2$ the antilinear mapping $$J(z) = \underbrace{\begin{bmatrix} 1 & i \\ 0 & 1 \end{bmatrix}}_{=:A} \bar z.$$ Then, we have $$J^2(z) = A\overline{(A \bar z)} = A\bar A z = \begin{bmatrix} 1 & -i + i \\ 0 & 1\end{bmatrix} z = z$$ and $$\langle J e_1, Je_2 \rangle = (A \bar e_1)^H (A \bar e_2) = (A e_1)^H (A e_2) = 1\cdot i + 0\cdot 1 = i \ne 0 = \langle e_2, e_1 \rangle.$$
1. Every antilinear mapping on $\mathbb C^n$ looks like $z\mapsto A\bar z$.
2. $A$ is symmetric(!) if and only if $J$ has the mentioned property: $$\langle J(u), J(v) \rangle = \langle A\bar u, A \bar v \rangle = \langle \bar u, A^H A \bar v \rangle = \langle \bar u, \bar v \rangle = \langle v, u \rangle$$ holds for every $u,v$ if and only if $A^H A = I$. That is $\bar A = A^{-1} = A^H = \bar A^T$. That is, $A$ is symmetric.
• sorry but what is this H notation. I mean $A^H$ – Sushil Mar 1 '16 at 19:30