Inner product equals for all vectors means Here, $\langle u,v\rangle$ denotes the inner product of $u$ and $v$.
Suppose $T$ is an operator in vectorspace $V$. Is it true that if $\langle Tv,v\rangle=\langle Tv,Tv\rangle$ for all $v\in V$ then $Tv=v$ for all $v$ in $V$? 
It seems obviously true but it seems like something that is not immediately trivial to prove... or maybe I'm missing something really simple.
 A: Your statement is not true.
Counterexample:
\begin{align}
T & = \begin{bmatrix}1&0\\0&0 \end{bmatrix},& 
v & = \begin{bmatrix} v_{1} \\ v_{2} \end{bmatrix} & \implies & &
T\,v = \begin{bmatrix} v_{1} \\ 0 \end{bmatrix}
\end{align}
\begin{align}
\left\langle T\,v ,\,v \right\rangle &  = \begin{bmatrix} v_{1} \\ 0 \end{bmatrix} \begin{bmatrix} v_{1} & v_{2} \end{bmatrix} = v_{1}^2 & 
\left\langle T\,v ,\,T\,v \right\rangle &  = \begin{bmatrix} v_{1} \\ 0 \end{bmatrix} \begin{bmatrix} v_{1} & 0 \end{bmatrix} = v_{1}^2
\end{align}
All you can say is that $\,T\,$ is orthogonal projection operator.
A: Counterexample 1. Let $T$ defined by $Tv=0$ for all $v\in V$, $T$ is an operator of $V$ such that:
$$\forall v\in V,\langle Tv,v\rangle=0=\langle Tv,Tv\rangle.$$
Nonetheless, if $V\neq\{0\}$, there exists $v\in V$ such that: $$Tv\neq v.$$
Counterexample 2. Let $W$ be a proper close subvector space of $V$, then one has : $$V=W\overset{\perp}{\oplus} W^{\perp}.$$
Let $p_{W}$ be the orthographic projection of $V$ on $W$ and of kernel $W^{\perp}$, notice that: $$\forall v\in V,p_W(v)\in W\textrm{ and }v-p_W(v)\in W^{\perp}.$$
Hence : $$\forall v\in V,\langle p_W(v),v-p_W(v)\rangle=0.$$
Therefore : $$\forall v\in V,\langle p_W(v),v\rangle=\langle p_W(v),p_W(v)\rangle.$$
Since $W\neq\{0\}$, $p_W\neq 0$ and since $W\neq V$, $p_W\neq\textrm{id}_V$.
A: Here's another failure mode; let the inner product be the dot product, and define
$$T = \left( \begin{matrix}\frac{1}{2} & \frac{1}{2} \\ -\frac{1}{2} & \frac{1}{2} \end{matrix} \right) $$
Writing $v = \left( \begin{matrix} x \\ y \end{matrix} \right)$, we have
$ Tv = \left( \begin{matrix} \frac{x+y}{2} \\ \frac{y-x}{2} \end{matrix} \right) $, and consequently,
$$ \langle Tv, v \rangle 
= \frac{1}{2} (x^2 + y^2)
= \langle Tv, Tv \rangle $$
The thing that makes this counterexample work is that
$$ T^t T = \left( \begin{matrix}\frac{1}{2} & 0 \\ 0 & \frac{1}{2} \end{matrix} \right) $$
and the fact that the expression $v^t A v$ depends only on the symmetric part of $A$; in the case of $2 \times 2$ matrices, it depends only on the top-left entry, the bottom-right entry, and the sum of the antidiagonal entries.
In general, the counterexamples are precisely the non-identity matrices $T$ satisfying $T^t T = \frac{T^t + T}{2}$.
