Null Space and Image of Projection I'm having trouble solving this question:
Consider in a vector space $V = M(3 × 3, \mathbb{C})$ of complex $n \times n$ matrices with inner product $\langle A, B\rangle = tr(A^∗B)$ the projection $P : V → V$
given by $P(X) = \frac{1}{2}(X + X^T)$. Give a basis for the null space and image of $P.$
1) For calculating the image I did $P(E_{11}), P(E_{12}), P(E_{13}), P(E_{21}), P(E_{22}), P(E_{23}), P(E_{31}), P(E_{32}), P(E_{33})$
where $E_{11}, E_{12}$... are the standard basis of V. 
My answer was a set of 6 matrices, 3 containing only the numbers 0 and 1, and 3 containing 0 and 1/2. The solution I was given to this was almost the same but using 1 where I had 1/2, so all matrices only contained zeros and ones. Why is that correct? Wouldn't I have to multiply all matrices by 2 in that case?
2) How can I calculate the basis of the null space? The solution is:
{$ \left( \begin{array}{cc}
0 & 1 & 0 \\
-1 & 0 & 0\\
0 & 0 & 0
\end{array} \right),
%
\left( \begin{array}{cc}
0 & 0 & 1 \\
0 & 0 & 0\\
-1 & 0 & 0
\end{array} \right),
%
\left( \begin{array}{cc}
0 & 0 & 0 \\
0 & 0 & 1\\
0 & -1 & 0
\end{array} \right)
$}
 A: There’s no unique solution to either part of this problem. Clearly, multiplying every element of a basis by some (perhaps different) nonzero scalar produces another basis for the same space. Compared to yours, the basis in the given solution has been “tidied up” so that the matrix entries are all integers, but that doesn’t make the solution any more correct than yours. 
For both parts of this problem, I recommend a “bigger picture” approach. By definition, the kernel of $P$ is the set of matrices for which $P(X)=\frac12(X+X^T)=0$, from which $X^T=-X$, as you noted in comments. Thus, the kernel of $P$ consists of all skew-symmetric matrices. Write down a generic element of this space and you should be able to read a basis for this space from it.  
Similarly, for the image of $P$ we can make use of the fact that $P^2=P$ for any projection, which in turn implies that its restriction to its image is the identity map. Therefore, the image of $P$ consists of the solutions to $\frac12(X+X^T)=X$, or $X^T=X$: the image is the space of symmetric matrices. As before, write down a generic element of this space and read a basis directly from it. With this approach, which is likely how the problem was meant to be solved, the resulting bases naturally end up with entries in $\{-1,0,1\}$. 
Note that for any square matrix $X$, $\frac12(X+X^T)$ is symmetric. Moreover every square matrix $X$ can be decomposed into the sum of a symmetric and skew-symmetric matrix as $\frac12(X+X^T) + \frac12(X-X^T)$. The first term is called the symmetric part of $X$.
