Here's a matrix problem: "Suppose A is idempotent. Let C=1/√2 * <

*|> (note: each <> identify a column in the matrix). Show that C is an involution.*- anonymous

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- anonymous

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- anonymous

If C is an involution C*C=I ¿right?

- anonymous

yep !

- anonymous

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- anonymous

Then, why don't follow this? C*C=1/√2 * <*| * 1/√2 * <**| =I, given that A is Idempotent, A*A=A*

- anonymous

The equallity should be fulfilled after perform all matrix operations

- anonymous

@satellite73 <*|> means |dw:1332812901190:dw|*

- anonymous

Yes, it you perform matrix operations you should verify the identity.

- anonymous

@juancarlosquintero_EC ok I tried it and I get C^2=1/2 * <*|<0, I+A>> but I don't get how is that supposed to be my identity matrix*

- anonymous

Could you type the course of your matrix operations, surely you didnt follow correctly rules for product of matrix.

- anonymous

In a sense, it looks similar to <<1, 0>|<0, 1>> which is the identity... do I need any numerical value ?

- anonymous

i think you can multiply block matrices easily
like regular matrix multiplication in a 2 by 2 matrix

- bahrom7893

WE ARE THE SATELLITES!

- anonymous

so you get a matrix that looks like
lol

- anonymous

\[
M =
\left[ {\begin{array}{cc}
II+AA & IA+AI \\
AI+IA & AA+II
\end{array} } \right]
\]

- anonymous

was it I or -I ?

- anonymous

it's -I in the second row, second column

- anonymous

ok i have to start again, but idea is clear right?

- anonymous

I guess... I'm writing what I did so you guys can see it

- anonymous

\[M =
\left[ {\begin{array}{cc}
II+AA & IA-AI \\
AI-IA & AA+II
\end{array} } \right]\]

- anonymous

this gives
\[M =
\left[ {\begin{array}{cc}
I+A & 0 \\
0 & A+I
\end{array} } \right]\]

- anonymous

yep this is what i got

- anonymous

with i guess a
\[\frac{1}{2}\] out front or something like that

- anonymous

yeppp

- anonymous

soo I+A and A+I are the pivots

- anonymous

now i guess if you square you should get the identity

- anonymous

ahhhhhhhh !!!

- anonymous

idempotent = projection, but i forget what an idempotent matrix looke like
mayby something like
\[
\left[ {\begin{array}{cc}
1 & 0 \\
0 &0
\end{array} } \right]\] although i don't think it matters for this problem

- anonymous

\[
\left[ {\begin{array}{cc}
I+A & 0 \\
0 & A+I
\end{array} } \right]^2\] can repeat as before since it is in blocks

- anonymous

and since it repeats... C is an involution

- anonymous

well now i am confused because i thought you were supposed to get the identity out of this. i don't see it

- anonymous

Me neither.

- anonymous

I only have to find the involution... If \[\left[\begin{matrix}I+A & 0 \\ 0 & A+I\end{matrix}\right]^{2}\] loops dosen't that mean it's an involution ? if yes then so is C, right ?

- anonymous

never mind this last comment

- anonymous

I switched involution and idempotent

- anonymous

ok thanks a lot guys for your help but honestly I'm very lost, I'll ask my teacher tomorrow...

- anonymous

actually i tried it with the simplest idempotent and identity i know, namely a two by two identity and a 2 by 2 idempototen
\[M =
\left[ {\begin{array}{cc}
1 & 0 \\
0 & 0
\end{array} } \right]\]

- anonymous

you can see the result here, but i if i divide by 2 it is still not the identity.
close though
http://www.wolframalpha.com/input/?i={{1%2C0%2C1%2C0}%2C{0%2C1%2C0%2C0}%2C{1%2C0%2C-1%2C0}%2C{0%2C0%2C0%2C-1}}^2

- anonymous

so actually i am not sure it is rigth

- anonymous

yeah not quite the identity but close.....

- anonymous

oh well you can ask tomorrow, and give this as a nice example where it does not quite work

- anonymous

XD ahahahahha I willl thanks a lot for your help it was very appreciated ! :P

- anonymous

@olibd :Did you ask your teacer about this?, What did he say about?

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