On
http://www.math.harvard.edu/~shlomo/docs/Advanced_Calculus.pdf
page 87, ex. 3.8, we read:
Given $V=M\oplus N$, prove that $V^*= M^0 \oplus N^0$, being $V^*$ the dual space of $V$ and $M^0$ the annihilator of M (same for $N$).
If $V$ is finite, I can see that $N$ is isomorphic to $M^0$ and vice-versa since $d(V)= d(M)+d(M^0)=d(N)+d(N^0)$ (but this is not exactly the proof, nor $V$ is assumed finite). Any hint?
Just use the definitions. If one desires to check that $V^* = M^0 \oplus N^0$, there are two things to be done.
1) $M^0 \cap N^0 = 0$. Let $f \in M^0 \cap N^0$. Then $f|_M = 0$ and $f|_N = 0$ gives $f|_{M \cup N} = 0$. But $f$ is linear, so it follows that $f|_{V}= f|_{M\oplus N} = f|_{{\rm span}(M\cup N)} = 0$.
2) $V^* = M^0 + N^0$. Let $f \in V^*$. We want to write $f = f_1+f_2$ with $f_1 \in M^0$ and $f_2 \in N^0$. Let's see what $f_1$ and $f_2$ must be. If an element decomposes as $v = m+n$, then we must have $$f(v) = f_1(n) + f_2(m).$$This motivates us to define $f_1 = f\circ \pi_N$ and $f_2 = f\circ \pi_M$, where $\pi_M$ and $\pi_N$ are the projections of $V$ onto $M$ and $N$, respectively. Clearly $f_1$ and $f_2$ are linear, with $f_1 \in M^0$ and $f_2 \in N^0$. And $$f_1(v) + f_2(v) = f(\pi_N(v)) + f(\pi_M(v)) = f(n)+f(m) = f(n+m) = f(v)$$as wanted.