Prove that $\operatorname{Trace}(A^2) \le 0$ Let $A \in M_n(\mathbb{R})$  is a antisymmetric matrix such as $A^T=-A$.
Prove that $\operatorname{Trace}(A^2) \le 0 $
I see that, for some matrix such as, their terms in diagonal are negative ?
 A: There are no doubt more sophisticated approaches, but every diagonal element of $A^2$ is obtained as $R_i\cdot C_i$ where $R_i$ and $C_i$ are row and column $i$ of $A$. Since $A$ is anti-symmetric one has $R_i=-C_i^T$, so $R_i\cdot C_i\leq0$ for each$~i$. With all diagonal elements of $A^2$ nonpositive, its trace certainly is so as well.
A: Multiply both sides of your equality by $A$, you get
$$AA^T=-A^2$$
but then $AA^T$ is positive definite, so $-AA^T=A^2$ is negative semi-definite, hence its trace is negative.
A: Another approach:
Let $e_i, \; 1\leq i \leq n$ be the standard basis of $\mathbb{R}^n$ (or any orthonormal basis), then 
$$\text{tr}(A^2) = \sum_{i=1}^n \langle A^2e_i, e_i \rangle = \sum_{i=1}^n \langle Ae_i, A^Te_i \rangle =-\sum_{i=1}^n \|Ae_i \|^2 \leq 0$$ 
A: Without revealing too much: rewrite $A^2$ using the equality that's been given. Then take a closer look at the diagonal elements, now that you have written the matrix square in a new way.
A: Let $\lambda$ be an eigenvalue of $A$.
Then using inner product we get
$$\langle Av, Av\rangle=\langle \lambda v, \lambda v\rangle=\bar{\lambda} \lambda.$$
On the other hand
$$\langle Av, Av\rangle =\bar{\lambda}\langle Av, v\rangle=
\bar{\lambda}\langle v, A^t v\rangle=\bar{\lambda}\langle v, -A v\rangle=
-\bar{\lambda} \bar{\lambda}\langle v,  v\rangle.$$
Therefore 
$$\bar{\lambda} \lambda=-\bar{\lambda}\bar{\lambda}.$$
Now either $\lambda=0$ or by dividing by $\bar{\lambda}$ we get $\lambda=-\bar{\lambda}$. Either way we get that all eigenvalues are with real part zero and the result follows.
A: Start with the Frobenius norm of A:
$$ \left\lVert A \right\rVert_F^2=  \mathrm{Tr}(A^TA)=\mathrm{Tr}(-AA)=\mathrm{Tr}(-A^2)=-\mathrm{Tr}(A^2)\geq 0 \Leftrightarrow \mathrm{Tr}(A^2) \leq 0 $$
