Is $M=\{A \in M^{\mathbb C}_{n \times n}\mid A=-\overline A\}$ a vector subspace over $\mathbb C$ and over $\mathbb R$? 
Is $M=\{A \in M^{\mathbb C}_{n \times n}\mid A=-\overline A\}$ a vector subspace over $\mathbb C$ and over $\mathbb R$ where for every $A=a_{ij}$ we define $\overline A=\overline a_{ij}$?

$\mathbf{1)M \quad over \quad \mathbb C}:$
As far as I understand:
if $a_{ij}=a+ib$ then $\overline a_{ij}=a-ib$ then $-\overline a_{ij}=-a+ib$
So essentially the condition is that $a=-a \Rightarrow a=0$.
$M$ is a subspace of all matrices of size ${n \times n}. $We can show that the zero vector is in $M$. 
Let $a_{ij}=a+ib, p_{ij}=c+id$.
$M$ is closed in terms of addition: 
$$
a_{ij}+p_{ij}=0+ib+0+id=0+i(b+d) \in M
$$
$M$ is closed in terms of scalar multiplication: 
$$
\lambda\cdot a_{ij}=\lambda\cdot (0+ib)=i\cdot \lambda b \in M
$$

$\mathbf{1)M \quad over \quad \mathbb R}:$
So in this case I'm not sure what happens. If $A=-\overline A$ and $a,b \in \mathbb R$ then:
1) $\overline a=a$ 
2) $\overline a=a \Rightarrow a=0$ 
So the elements of the matrices can be only zeroes. Which of course would be a vector subspace but this is strange.
 A: The set $M$ is closed under addition: if $A,B\in M$, then
$$
-\overline{A+B}=-\bar{A}-\bar{B}=A+B
$$
so you only have to check closure under product with scalars.
Note that, if $a\in\mathbb{C}$ and $A\in M$,
$$
-\overline{aA}=-\bar{a}\bar{A}=\bar{a}(-\bar{A})=\bar{a}A
$$
so
$$
aA\in M\quad\text{if and only if}\quad aA=\bar{a}A
$$
This surely happens if $a\in\mathbb{R}$. Can you say it always happens when $a\in\mathbb{C}$, for all $A\in M$?
A: You seem to have misunderstood the question.  In both parts, we're conidering the set $M=\{A \in M^{\mathbb C}_{n \times n}|A=-\overline A\}$, which is a set of matrices with complex entries.
A set a subspace over $\Bbb C$ if it is closed under addition and multiplication by complex scalars. A set a subspace over $\Bbb R$ if it is closed under addition and multiplication by real scalars.  Since the set $M$ is closed under addition, it suffices to consider the condition of closure under scalar multiplication.
In other words, checking whether $M$ is a vector space over $\Bbb C$ means answering the question: for $\lambda = a + bi$, is it true that $A \in M \implies \lambda A \in M$? (The answer should be no). Checking whether $M$ is a vector space over $\Bbb R$ means answering the question: for $\lambda \in \Bbb R$, is it true that $A \in M \implies \lambda A \in M$? (The answer should be yes).  
A: Using your approach. 
You have noted that the entries of the matrices $A$  are pure imaginary numbers (and the sum of pure imaginary numbers is pure imaginary) so for the scalar multiplication we have:
$$
\lambda (0+ib)=i (\lambda b)
$$ 
clearly  ( for $b\ne0$) this is pure imaginary only if $\lambda$ is a real number. 
So $M$ is a vector space over $\mathbb{R}$ but bot over $\mathbb{C}$. 
Note that to be a vector space over $\mathbb{R}$  does not means that the entries of the matrices are real numbers, but that the field of scalars is $\mathbb{R}$ .This is the cause of your added doubt.
