Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
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
up vote 2 down vote accepted

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.

share|cite|improve this answer
tried to redo the problem without glimpsong at your solution and I'm very happy to see that I got the same solution as yours :) Thanks for the confirmation! – sigmatau Jan 13 '14 at 20:13

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.