The following theorem is true:
Theorem: Let $V$ be a finite dimensional unitary vector space und $\phi: V \rightarrow V$ a normal, linear transformation with $\langle v, \phi(v) \rangle = 0$ for all $v \in V$. Then $\phi = 0$.
Proof: Since $\phi$ is normal and $V$ is finite dimensional and unitary, by the spectral theorem, there exists an Eigenbasis $B$ of $V$ (which is orthonormal). Hence we may write the transformation matrix $D$ of $\phi$ as $\text{diag}(\lambda_1, \ldots, \lambda_n)$ with $\lambda_i \in \mathbb{C}$ for all $i \in \{1, \ldots, n\}$. Hence for $v \in V$
$$ \langle v, \phi(v) \rangle = \langle v, Dv\rangle = \sum_{i = 1}^n \lambda_i \vert v_i \vert^2 = 0$$
This is then in particular true for $e_i = (0, \ldots, 1, \ldots, 0)$ with the $1$ in the $i$-th place. Hence for every $i \in \{1, \ldots, n\}: \langle e_i, \phi(e_i) \rangle = \lambda_i \vert e_i \vert^2 = \lambda_i = 0$. Hence $D$ is the $0$-matrix on $B$. Thus $\phi$ and $0$ agree on the basis $B$ and thus agree as lineare transformations. QED.
Now, aparently, this is not true for euclidean spaces, but I cannot spot the mistake when replacing unitary with euklidean. A counter example would be appreciated just as well.
Thank you in advance.
Edit:
The proof I have given is false (or at least not formal/precise). Here is a better one, which uses the same idea though:
Since $\phi$ is normal and $V$ is finite dimensional and unitary, by the spectral theorem, there exists an eigenbasis $B:= \{b_1, \ldots, b_n\}$ of $V$ (which is orthonormal). Write $\phi(b_i) = \lambda_i b_i$ with $\lambda_i \in \mathbb{C}$ the associated eigenvalue to $b_i$. Hence in particular for the $b_i$ we have
$$ \langle b_i, \phi(b_i) \rangle = \langle b_i, \lambda_i b_i \rangle = \lambda_i \langle b_i, b_i \rangle = \lambda_i = 0$$
Thus $\phi$ and $0$ agree on the basis $B$ and thus agree as lineare transformations. QED.
A counterexample is the rotation by $\frac{\pi}{2}$ in the standard euclidean plane. Its matrix is $\left(\matrix{0&-1\\1&0}\right)$.
Its adjoint (that is, its transpose in this case) is also its inverse, hence they commute and the operator is normal as required. As for the $\left<v,\varphi(v)\right>=0$ condition, it follows from the geometric interpretation of $\varphi$ as a rotation by $\frac{\pi}2$.