Showing a Linear Transformation is $0$ through an Inner Product Condition $\fbox{Setting}$
Let $V$ be an inner product space and $\tau$ a linear transformation on $V$.
Suppose that $\exists w \in V$ s.t. $\forall u \in V$, $\left\langle \tau(u),w \right \rangle = 0$.
$\fbox{Claim}$ $\tau = 0$.
$\fbox{Argument}$


*

*Pick any $u \in V$.

*Then $\left\langle \tau(u), w\right\rangle = 0 = \left\langle0,w\right\rangle$ so that $\tau(u) = 0$.

*Doing this for all $u \in V$ shows that $\tau = 0$.
Does this work?  If it does work, couldn't we use similar reasoning to claim absurdly that $w = 0$ via $\left\langle \tau(u),w\right\rangle = 0 = \left\langle \tau(u),0\right\rangle$?
EDIT: Per the answers below -- is there a way that the above hypothesis could be naturally modified s.t. the conclusion actually holds?
 A: This claim is untrue. If $\tau$ projects any vector into the space orthogonal to $w$, then the inner product will always be zero with $w$ by construction. But this projector is not zero.
The second step of your argument is incorrect.
A: Your error lies in the statement
$$\left\langle \tau(u), w\right\rangle = 0 = \left\langle0,w\right\rangle$$
Two non-zero vectors may give an inner product that is zero, so there is no reason to believe that $\tau (u) = 0 $ is true.
For a counter example to the original claim, consider the projection onto a line, and use $w$ orthogonal to the line.
A: The claim is not true! In fact, every linear transformation $\tau$ has this property. We simply take $w=0$; we know $0 \in V$ and $\tau(0) = 0$ (since $\tau$ is a linear transformation), so this works. (And of course, not every linear transformation $\tau$ has $\tau \equiv 0$)
As the other answer says, the claim is still untrue even if we can find a non-zero $w$.
Remember, 
$$\langle x, y \rangle = 0$$
only tells you that $x$ and $y$ are orthogonal. It cannot tell you that either is 0!
