Sesquilinear form with an operator

Question

Suppose I have an operator $T$ such that $$\forall v\in V,\langle v|Tv\rangle=\langle Tv|v\rangle$$ (where the inner product is defined on complex rather than real)

Will it necessarily imply that $$\langle v|Tw\rangle=\langle Tv|w\rangle \forall v,w\in V?$$

I thought about computing $\langle(v+w)|T(v+w)\rangle$ and $\langle T(v+w)|v+w\rangle$ since they are equal but I got an equality which is not at all conclusive.

Answer 1

The current answers don't properly cover complex vector spaces, where the components of vectors can be complex, and the inner product is sesquilinear, rather than bilinear.

For complex vector spaces the result will turn out to be true. We first prove a polarisation identity: Let $V$ be a complex vector space, and $v,w \in V$. If we write $t(v,w) = \langle v | T w \rangle$, and $N(v) = t(v,v)$, $t$ is a sesquilinear map on $V$ (since $T$ is linear and the inner product is sesquilinear), and $N$ satisfies $$ N(v+w) = N(v) + N(w) + t(v,w) + t(w,v) $$ and $N(\alpha v) = \lvert \alpha \rvert^2 N(v)$ for any $\alpha \in \mathbb{C}$ and $v,w \in V$. Replacing $w$ by $ i^n w $ gives $$ N(v+i^k w) = N(v) + N(w) + i^k t(v,w) + i^{-k} t(w,v) $$ using the sesquilinearity, so $$ \sum_{k=0}^3 i^{-k} N(v+i^k w) = 4 t(v,w) $$ (this argument works for absolutely any sesquilinear map).

What about the condition in the question? Since the inner product satisfies $\langle v | w \rangle = \overline{\langle w | v \rangle}$ for every $v,w \in V$, the condition can be written $ N(v) = \langle v | Tv \rangle = \langle Tv | v \rangle = \overline{\langle v | Tv \rangle} = \overline{N(v)} $, i.e. $N(v)$ is real for every $v$. But this means that $$ \begin{align} \langle Tv | w \rangle &= \overline{\langle w | Tv \rangle} \\ &= \overline{t(w,v)} \\ &= \frac{1}{4} \sum_{k=0}^3 \overline{i^{-k} N(v+i^k w)} \\ &= \frac{1}{4} \sum_{k=0}^3 i^k N(v+i^k w) \end{align} $$ But $ N(v+i^k w) = N(i^k(w+i^{-k}v)) = N(w+i^{-k}v) $ by sesquilinearity again, so this is equal to $$ \frac{1}{4} \sum_{k=0}^3 i^k N(w+i^{-k} v) = \frac{1}{4} \sum_{r=0}^3 i^{-r} N(w+i^r v) = t(v,w) = \langle v | Tw \rangle , $$ as required.

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This doesn't work for real spaces because the inner product is actually symmetric there (i.e. $\langle v | w \rangle = \langle w | v \rangle$ for every $v,w$), so every linear map satisfies the condition in the question; on the other hand, the complex case does give extra information, namely the reality of $N(v)$.

Answer 2

NO.

Consider $$ T=\left(\begin{array}{cc} 1&1\\ 0& 1\end{array}\right) $$ Then, for $v=(v_1,v_2)$, $$ \langle Tv,v\rangle=v_1^2+v_1v_2,v_2^2=\langle v,Tv\rangle $$ but for $u=(u_1,u_2)$, $v=(v_1,v_2)$ $$ u_1^2+u_2v_1+u_2v^2=\langle Tu,v\rangle\ne \langle u,Tv\rangle=u_1^2+u_1v_2+u_2v^2. $$

Answer 3

N0. Take $T(x,y)=-(y,x)$. Than you always have $\bigl\langle T(v)\mid v\bigr\rangle=\bigl\langle v\mid T(v)\bigr\rangle=0$. But you don't always have $\bigl\langle T(v)\mid w\bigr\rangle=\bigl\langle v\mid T(w)\bigr\rangle$.