Proving facts about skew symmetric matrices using a non-standard definition A linear transformation $A:\mathbb{R}^n\rightarrow\mathbb{R}^n$ is skew symmetric if $x\cdot(Ay)=-(Ax)\cdot y$ for every pair of vectors $x,y\in\mathbb{R}^n$. Assume that $n=3$ and $A$ is skew symmetric.
(a) Show that $0$ must be an eigenvalue of $A$.
(b) Show that the diagonal entries of the matrix representation of $A$ with respect to any orthonormal basis $\{e_1,e_2,e_3\}$ are all zero.
The traditional definition of a skew symmetric matrix that I've seen is that $A^T=-A$. Using this definition, solving (a) is just a matter of manipulating properties of determinants, and solving (b) comes from a straightforward element argument ($A_{ii}=-A_{ii}\Rightarrow A_{ii}=0$). I've been trying to relate this definition of skew symmetric to the one I've been given, but I'm not seeing the correspondence.
 A: To show $0$ is an eigenvalue take $x=y=e_1+e_2$ in the definition
To show the second proposition take $x=y=e_1$.
A: $\det(A-\lambda I_3)$ is a cubic polynomial, so it must have a real root, i.e. it must have an eigenvalue. Now, assume that $(\lambda, v)$ is an eigenpair of $A$.
Then:
$$\begin{array}{rcl}
v \cdot (Av) &=& -(Av) \cdot v \\
v \cdot (\lambda v) &=& -(\lambda v) \cdot v \\ 
\lambda (v \cdot v) &=& -\lambda (v \cdot v) \\ 
2 \lambda (v \cdot v) &=& 0 \\
\lambda &=& 0
\end{array}$$
because $v \cdot v \ne 0$, because $v \ne 0$.

Note that the $ij$-th entry of the matrix is $e_i^T A e_j$, i.e. $e_i \cdot (Ae_j)$ (I may have reversed the order).
Now, for every $i \in \{1,2,3\}$, the $ii$-th entry is $e_i \cdot (Ae_i)$, and the same argument applies:
$$\begin{array}{rcl}
e_i \cdot (A e_i) &=& -(A e_i) \cdot e_i \\
2 e_i \cdot (A e_i) &=& 0 \\
e_i \cdot (A e_i) &=& 0
\end{array}$$
So the $ii$-th entry is $0$.
A: You can actually prove that every matrix representation of $A$ with respect to any orthonormal basis $e = \{e_1, e_2, e_3\}$ will satisfy $A(e)^T = -A(e)$.
Indeed, let $A(e) = (a_{ij})$. We have:
$$a_{ij} = \langle e_i, Ae_j \rangle = - \langle Ae_i, e_j\rangle = -a_{ji}$$
Therefore $A(e) = -A(e)^T$.
$(b)$ immediately follows because $a_{ii} = -a_{ii} \implies a_{ii} = 0$.
For $(a)$, notice that:
$$\det A(e) = \det(-A(e)) = (-1)^3 \det A(e) = -\det A(e) \implies \det A(e) = 0$$
Therefore, $\det A = 0$ so $0$ is an eigenvalue of $A$.
