Is $ f(A) = A + 2A^{T} $ an isomorphism of $ \mathbb R^{5,5} $ onto itself? I have problem with prove or disprove this hypothesis:

Is the linear transformation $ f \in L(\mathbb R^{5,5},\mathbb R^{5,5}) $
$$ f(A) = A + 2A^{T} $$
  an isomorphism of the space $ \mathbb R^{5,5} $ onto itself?

In order to be an isomorphism, $f$ must be injective and surjective.
Checking if it is surjective:
Assume that as a result we want to get:
$$\mathbf{Q} = 
 \begin{bmatrix}
 q_{11} & q_{12} & \cdots & q_{1n} \\
 q_{21} & q_{22} & \cdots & q_{2n} \\
 \vdots & \vdots & \ddots & \vdots \\
 q_{m1} & q_{m2} & \cdots & q_{mn}
 \end{bmatrix} $$
and we put as argument $$  \mathbf{A} = 
 \begin{bmatrix}
 a_{11} & a_{12} & \cdots & a_{1n} \\
 a_{21} & a_{22} & \cdots & a_{2n} \\
 \vdots & \vdots & \ddots & \vdots \\
 a_{m1} & a_{m2} & \cdots & a_{mn}
 \end{bmatrix} $$
then choose $i,j$. Factors $a_{ij}$ $a_{ji}$ $q_{ij}$ $q_{ji}$ are only in this linear system:
$$q_{ij} = a_{ij} + 2a_{ji}  \wedge  q_{ji} = a_{ji} + 2a_{ij}$$
so $$ a_{ij} = \frac{2q_{ij}-q_{ji}}{3} $$ and $$ a_{ji} = \frac{2q_{ji}-q_{ij}}{3} $$
so the factors $ a_{ij}$ $ a_{ji}$ are determined unambiguously. So it is surjective.
But how to deal with checking injectivity?
Suppose that $$ A+2A^T=B+2B^T $$
$$ A-B = 2(A^T-B^T) $$
and I don't know how to finish that.
 A: Hint:
Check if the kernel of this transformation is trivial. This is enough to establish if it is isomorphism: 
$$\begin{eqnarray}
f(A) =0  
&\implies &A=-2A^T \\
&\implies &A^T=(-2A^T)^T \\
&\implies &A^T = -2A \\
&\implies &A =4A \\
&\implies &A=0
\end{eqnarray}
$$
A: You have $A-B = 2(A-B)^\top$.
This implies $a_{ij} - b_{ij} = 2 (a_{ji} - b_{ji}) = 4(a_{ij} - b_{ij})$, so...
A: You're already done. Since the map is from $\mathbb R^{5\times 5}$ to $\mathbb R^{5\times 5}$, checking that it's surjective (that it has full rank) also tells you that it's injective (that it has nullity $0$) by the rank-nullity theorem.
A: Hint:
It suffices to check that $f$ is surjective. For any matrix $B$ we have
$$f\left(-\frac13B + \frac23B^T\right) = B$$

It you are wondering how we got this, assume $A+2A^T = B$. Then $A^T + 2A = B^T$ so $$B + B^T = 3(A+A^T)$$
Now it follows $$A = B^T - (A+A^T) = B^T - \frac13(B + B^T) = -\frac13B + \frac23B^T$$
A: When you say
$$
a_{ij}=\frac{2q_{ji}-q_{ij}}{3}
$$
(I fixed the indices) is “determined unambiguously”, you have proved both surjectivity and injectivity. There is at most one matrix $\mathbf{A}$ such that $f(\mathbf{A})=\mathbf{Q}$. And there is one, determined in the way you did (after fixing the small mistake).

On the other hand, a linear map $L\colon V\to V$, where $V$ is a finite dimensional vector space, is injective if and only if it is surjective, because of the rank-nullity theorem. Therefore just one of the properties needs to be checked for.
You don't need to go to the level of matrix entries in order to prove surjectivity. Given $Q$ you want to find $A$ such that $f(A)=Q$, that is, $A+2A^T=Q$. If such matrix exists, then also
$$
A^T+2A=Q^T
$$
and so $2A^T+4A=2Q^T$; thus
$$
Q-2Q^T=A+2A^T-2A^T-4A=-3A
$$
so
$$
A=\frac{1}{3}(2Q^{T}-Q)
$$
is the only possible one. Since
$$
f\left(\frac{1}{3}(2Q^{T}-Q)\right)=Q
$$
as it can be readily checked, you have surjectivity and injectivity.
Checking directly for injectivity is easier, though.
