In the space C^2 of all vectors (a1, a2)
Let M = {(a1, a2) such that a1=a2}
N1 = {(0, a1)}
N2 = {(a1, 0)}
If E1and E2 are projections on M along N1, N2, respectively, show that E1E2 = E2 and E2E1 =E1
Please, help

Question

In the space C^2 of all vectors (a1, a2) 
Let M = {(a1, a2) such that a1=a2}
      N1 = {(0, a1)} 
      N2 = {(a1, 0)}
If E1and E2 are projections on M along N1, N2, respectively, show that E1E2 = E2 and E2E1 =E1
Please, help

loser66 · Answer

all elites I know are here. Help me, friends :)

kinggeorge · Answer

I'm a little confused as to what E1 and E2 do. Do they project a vector in M onto N1 and N2? Or do they take a vector in N1 or N2 and project it onto M?

loser66 · Answer

I think E1 sends a vector in N1 on M. and the same with E2

kinggeorge · Answer

I'm a bit confused, since it seems like if we apply E1 to a vector in N1, then we should get something that isn't in N2, and so we can't apply E2 afterwards. But then E2E1 would be undefined... Unless the projections just send everything to 0.

loser66 · Answer

Let me type the definition of the projection:
If V is the direct sum of M and N, so that every z in V may be written, uniquely, in the form of z = x + y, with x in M and y in N, the projection on M along N is the transformation E defined by  Ez =x
 from it, you can see, "on M along N" and the result is "x", 
Apply to my problem "on M along N1" mean the target set is M

kinggeorge · Answer

It seems to me as if this projection is defined only if $M=N_1\oplus N_2$. But in this example, that's not the case. $M$ is merely a subspace of $N_1\oplus N_2$. So I'm not really sure how to proceed.

loser66 · Answer

let say C^2 = N1directsum N2, so, M is subspace of C^2

kinggeorge · Answer

So when I'm reading over this definition, it only makes sense to me when we have two spaces$$V_1=M\oplus N_1$$$$V_2=M\oplus N_2$$Then for a vector $v+w\in M\oplus N_1$, the projection on M along $N_1$ is what I would expect it to be, that is $E_1(v+w)=v$.

kinggeorge · Answer

So if let a general vector in $M$ be $(a_1,a_1)$, and $(0,a_2)$ a general vector in $N_1$, we have$$E_1((a_1,a_1)+(0,a_2))=(a_1,a_1).$$But this vector is also in $V_2$ as $(a_1,a_1)+(0,0)$. So$$E_2((a_1,a_1))=(a_1,a_1).$$So if $v\in V_1$, we have $E_2(E_1(v))=E_1(v)$ which is what we wanted to show.

loser66 · Answer

mistypo? N1 ={(0,a1)} not (0,a2)

loser66 · Answer

oh, I am sorry, you are right, (0,a2) in N1

kinggeorge · Answer

The $a_1$ for $N_1$ should be different that the $a_1$ in $M$. So I specifically made it $a_2$ so there wouldn't be a risk of confusing what vector was from what.

kinggeorge · Answer

A very similar argument would work for showing that $E_1(E_2(v))=E_2(v)$ if $v\in V_2$.

loser66 · Answer

I got what you mean, thanks a lot, One more question need your brain, Please!

kinggeorge · Answer

I'll see what I can do.

loser66 · Answer

the same but L ={(a1, -a1)}
if E0 is the projections on N2 along L, then E2E0 is a projection but E0E2 is not

kinggeorge · Answer

So if we have $v=(a_1,a_2)\in\mathbb{C}^2$, then we can say that $(a_1,a_2)=(-a_2,a_2)+(a_2+a_1,0)\in L\oplus N_2$. Then$$E_0(v)=(a_1+a_2,0)$$In $M\oplus N_2$, this vector is $(0,0)+(a_1+a_2,0)$, so that$$E_2((a_1+a_2,0))=(0,0).$$

kinggeorge · Answer

If we do it the other way, then $(a_1,a_2)$ can be written as $(a_2,a_2)+(a_1-a_2,0)$ in $V_2$. Applying $E_2$ to this, we get$$E_2(a_1,a_2)=(a_2,a_2).$$This vector is just $(-a_2,a_2)+(2a_2,0)$ in $L\oplus N_2$. So applying $E_0$ to this,$$E_0((a_2,a_2))=(2a_2,0).$$

loser66 · Answer

For part a, is it E1E2 = E2E1= E1 = E2? After calculate, I got it. not sure!!

kinggeorge · Answer

I don't think that's necessarily the case since the vectors will have different decompositions depending on whether you're considering it as part of $M\oplus N_1$ or $M\oplus N_2$. If they DO have the same decomposition I think it might be true, but that's not always the case.

kinggeorge · Answer

A general vector $(a_1,a_2)\in\mathbb{C}^2$ will be decomposed as $(a_1,a_1)+(0,a_2-a_1)$ in $M\oplus N_1$ and $(a_2,a_2)+(a_1-a_2,0)$ in $M\oplus N_2$.

kinggeorge · Answer

So $E_1E_2(v)=(a_2,a_2)$, but $E_2E_1(v)=(a_1,a_1)$.

loser66 · Answer

Give me your book, please!! how can you see it clearly like that?

kinggeorge · Answer

I was using the following site for a formal definition of this projections, and after that it's just a matter of practicing the stuff. But to see that $E_1E_2(v)=(a_2,a_2)$ and $E_2E_1(v)=(a_1,a_1)$, I just looked at the computations I did near the top. http://en.wikibooks.org/wiki/Linear_Algebra/Projection_Onto_a_Subspace

loser66 · Answer

Thanks a ton.

kinggeorge · Answer

You're welcome. I enjoy thinking about the slightly more abstract questions on here rather than the mere computational questions such as "find this integral" or "what are the roots of this polynomial"

loser66 · Answer

I work on it, as you said E1((a1, a1) + (0, a2) ) = (a1, a1)
and the result is on V2, too, so that E2(a1,a1) = (a1,a1) $ {\color{red}{question~~is~~ here: is~~it~~ E2((a1,a1) +(a1,0)) = (a1,a1)}}$
and how to express E2E1 in terms of E1 to get E2E1 = E1? 
Since you said (a1, a1) is in V2 "But this vector is also in V2 as (a1,a1)+(0,0)"
I don't see that vector is in V2 because of (0,0) is not in N2
Please, explain me

loser66 · Answer

@KingGeorge

kinggeorge · Answer

$(0,0)$ is indeed in $N_2$. You have $(0,0)=(a_1,0)$ where $a_1=0$.

loser66 · Answer

oh!! so, if they say (a1,0) we can have the case a1 =0 hihihi.... stupid me.
Ok, I got that part
how about E2E1 = E1, I confused

kinggeorge · Answer

To show that $E_2E_1=E_1$, we need to show that for any vector $v\in\mathbb{C}^2$, $E_2(E_1(v))=E_1(v)$. So if $v=(a_1,a_2)$, then $v=(a_1,a_1)+(0,a_2-a_1)$ in $M\oplus N_1$, so that$$E_1(v)=(a_1,a_1).$$Now when we consider this vector as part of $M\oplus N_2$, we see that $v=(a_1,a_1)+(0,0)$, so that$$E_2(E_1(v))=E_2(a_1,a_1)=(a_1,a_1).$$So $E_2(E_1(v))=E_1(v)$ for any choice of $v\in\mathbb{C}^2$, so $E_2E_1=E_1$.

loser66 · Answer

YYYYYYYYYYYes, hihihi, Thank you.

kinggeorge · Answer

No problem.