anonymous
  • anonymous
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.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
If C is an involution C*C=I ¿right?
anonymous
  • anonymous
yep !
anonymous
  • anonymous
but what does <|> mean? because my guess isi if you square it and use the fact that \[A^2=A\] it should pop right out

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anonymous
  • anonymous
Then, why don't follow this? C*C=1/√2 * <| * 1/√2 * <| =I, given that A is Idempotent, A*A=A
anonymous
  • anonymous
The equallity should be fulfilled after perform all matrix operations
anonymous
  • anonymous
@satellite73 <|> means |dw:1332812901190:dw|
anonymous
  • anonymous
Yes, it you perform matrix operations you should verify the identity.
anonymous
  • 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
  • anonymous
Could you type the course of your matrix operations, surely you didnt follow correctly rules for product of matrix.
anonymous
  • anonymous
In a sense, it looks similar to <<1, 0>|<0, 1>> which is the identity... do I need any numerical value ?
anonymous
  • anonymous
i think you can multiply block matrices easily like regular matrix multiplication in a 2 by 2 matrix
bahrom7893
  • bahrom7893
WE ARE THE SATELLITES!
anonymous
  • anonymous
so you get a matrix that looks like lol
anonymous
  • anonymous
\[ M = \left[ {\begin{array}{cc} II+AA & IA+AI \\ AI+IA & AA+II \end{array} } \right] \]
anonymous
  • anonymous
was it I or -I ?
anonymous
  • anonymous
it's -I in the second row, second column
anonymous
  • anonymous
ok i have to start again, but idea is clear right?
anonymous
  • anonymous
I guess... I'm writing what I did so you guys can see it
anonymous
  • anonymous
\[M = \left[ {\begin{array}{cc} II+AA & IA-AI \\ AI-IA & AA+II \end{array} } \right]\]
anonymous
  • anonymous
this gives \[M = \left[ {\begin{array}{cc} I+A & 0 \\ 0 & A+I \end{array} } \right]\]
anonymous
  • anonymous
yep this is what i got
anonymous
  • anonymous
with i guess a \[\frac{1}{2}\] out front or something like that
anonymous
  • anonymous
yeppp
anonymous
  • anonymous
soo I+A and A+I are the pivots
anonymous
  • anonymous
now i guess if you square you should get the identity
anonymous
  • anonymous
ahhhhhhhh !!!
anonymous
  • 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
  • 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
  • anonymous
and since it repeats... C is an involution
anonymous
  • 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
  • anonymous
Me neither.
anonymous
  • 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
  • anonymous
never mind this last comment
anonymous
  • anonymous
I switched involution and idempotent
anonymous
  • anonymous
ok thanks a lot guys for your help but honestly I'm very lost, I'll ask my teacher tomorrow...
anonymous
  • 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
  • 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
  • anonymous
so actually i am not sure it is rigth
anonymous
  • anonymous
yeah not quite the identity but close.....
anonymous
  • anonymous
oh well you can ask tomorrow, and give this as a nice example where it does not quite work
anonymous
  • anonymous
XD ahahahahha I willl thanks a lot for your help it was very appreciated ! :P
anonymous
  • anonymous
@olibd :Did you ask your teacer about this?, What did he say about?

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