Skew matrix properties 
Let $f$ be a linear transformation on the euclidean space $E$. Prove that the following statements are equivalents

*

*there is a orthonormal basis where the matrix representation of $f$ is skew symmetric,

*the matrix representation of $f$ with respect to any orthonormal basis is skew symmetric,

*$\forall x\in E, \langle x,f(x)\rangle=0$.


By showing that every statements are equivalent to $f$ being skew symmetric  ($\langle x,f(y)\rangle=-\langle f(x),y\rangle$), it is easy to conclude that they are equivalent but I feel like the way the exercise was asked there is another way to solve it but I don't see how.
 A: Typically such problems are intended to be solved by finding a circle of implications. In this case, we can prove $3\implies 2\implies 1\implies 3$.
First note that if $A=[f]_{e_i}$ is the matrix representation of $f$ w.r.t. an orthonormal basis $e_i$, $A_{ij}=\langle e_i,f e_j\rangle$. 
Now we prove $3\implies 2$. Let $e_i$ be any orthonormal basis, we want to show given 3 that $\newcommand{\ang}[1]{\left\langle{#1}\right\rangle}A_{ij} = -A_{ji}$. By the above note and 3, we have 
$$A_{ij}+A_{ji} + 0 + 0=\ang{e_i,f e_j}+\ang{e_j,f e_i} + \ang{e_i,f e_i} + \ang{e_j, f e_j} = \ang{e_i + e_j, f (e_i+e_j)} = 0,$$
so 2 holds.
That 2 implies 1 is obvious, so we just need to prove 1 implies 3.
Now note that $T:x\mapsto \ang{x,f x}$ is a linear map, so to check whether or not it's zero, it suffices to check whether it's zero on a basis. In particular, suppose $[f]_{e_i}$ is skew symmetric for some orthonormal basis $e_i$. Then $0=A_{ii}=\ang{e_i,f e_i}=Te_i$. Thus $T$ is zero on a basis, hence zero everywhere, so 3 holds.
