uniqueness of map proof

Let $W_1,W_2 \in V$ be complementary subspaces. Show that there exists a unique map $P$ with with Kern$P=W_2$ and Im$P=W_1$ such that $P=P^2$

I prooved the existence of such a map and it is in accordance with the proof presented in my solutions (this is a homework). But I have a different proof of uniqueness and am not sure of it's correctness.It goes as follows.

Let $P$ and $F$ be two maps with the above property. Let $v_1,v_2,....v_s$ and $v_{s+1},...,v_n$ be the Bases of $W_1$ and $W_2$ respectivly. We define $P(v_i)=v_i=F(v_i)$ for $1 \leq i \leq n$. Then we have for arbitrary $v \in V, v:=\lambda_1 v_1+...+\lambda_nv_n$ that $P(v)=P(\lambda_1 v_1+...+\lambda_nv_n)=\lambda_1P(v_1)+..+\lambda_sP(v_s)+\lambda_{s+1}\cdot 0+...+\lambda_n \cdot 0=\lambda_1P(v_1)+...+\lambda_sP(v_s)=\lambda_1 F(v_1)+...+\lambda_sF(v_s)=F(\lambda_1v_1+...+\lambda_nv_n)=F(v)$.

Hence $F$ and $P$ must be equal.

Is this a correct approach?

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The maps $P$ and $F$ are given, you cannot "define $P(v_i)=v_i=F(v_i)$", doing so you're just assuming $P$ and $F$ coincide in $W_1$. – Vinicius M. Jan 12 '14 at 19:59
@HagenvonEitzen fixed it. – sigmatau Jan 12 '14 at 20:50

The question asks you to show uniqueness of a map $P$ satisfying $P^2 = P$ and $\ker P = W_2$, $\text{im } P = W_1$, so what you need to show is that if there is another map $F$ with $\ker F = W_2$, $\text{im } F = W_1$, and $F^2 = F$, then $P = F$. The reasoning you have given is not correct: you cannot simply define $P$ and $F$, you can only assume they have the same kernels and images (and are both idempotent).
A direct way to show $P = F$ would be the following: any $v \in V$ can be written as $v = v_1 + v_2$, where $v_1 \in W_1, v_2 \in W_2$. Since $\ker P = W_2 = \ker F$, $P(v_2) = 0 = F(v_2)$. Also, since $\text{im } P = W_1 = \text{im } F$, we can write $P(w) = v_1 = F(w')$ for some vectors $w, w'$. Then $P(v) = P(v_1 + v_2) = P(v_1) = P(P(w)) = P(w) = v_1$, and by the exact same reasoning, $F(v) = F(v_1 + v_2) = F(v_1) = F(F(w')) = F(w') = v_1$ as well.