For finite dimensional vector spaces show that $(V \times W)' \cong V' \times W'$ To show: 
For finite dimensional vector spaces $V, W$ show that $(V \times W)' \cong V' \times W'$.
There is a hint : If $f \in V' $ and $g \in W'$ there exists a natural isomorphism defined as follows: let $T(f,g)$ be the linear form $(x,y) \mapsto f(x) + g(y) $ on $(V \times W)$. I want to use the hint. 
Edit: the prime notation is Sterling Berberian's notation for the dual space.
Thoughts: First since $f$ and $g$ are linear, then the linearity of $T$ follows.
Surjectivity, Take any $(f,g) \in (V \times W)' $ then $ T(f,g) = f(x) + g(y) = (f,g)(x,y)$.
Injectivity: Not sure how to proceed.
 A: The claim follows immediately by counting dimensions, because $$\dim (V \times W)'=\dim(V \times W)=\dim V + \dim W=\dim(V' \times W').$$ Of course, you would like a 'natural' isomorphism. The hint already gives one, namely $T:V'\times W' \rightarrow (V \times W)'$. It is easily checked $T$ is linear. Now we only need to check injectivity, because then, the image of $T$ has the dimension of its domain, which is the same as the dimension of its codomain(here we use the fact the dimensions of $V$ and $W$ are finite!). So, suppose $T(f,g)=0$ for some linear functionals $f \in V'$, $g \in W'$. Then $f(x)+g(y)=0$ for all choices of $(x,y) \in V \times W$. So in particular $f(x)=0$ for all $x \in V$, so $f=0$. Similarly, $g=0$, so $(f,g)$ is the zero vector in $V'$, and $\ker T$ is trivial, so $T$ is injective. 
A: Suppose $T(f,g) = 0$, then $f(x) + g(y) = 0$ for all $(x,y) \in (V \times W)$ 
thus $f(x) = -g(y)$ for all $(x,y) \in (V \times W)$. Take any $(x,0) \in (V \times W)$  then $f(x) = -g(0) = 0$ for all $x \in V$. Then $f$ must be the zero function and similarly for $g(y)$. Then $f=g=0$ ( the zero function) and T is injective.
