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anonymous
 4 years ago
Help Please :D
anonymous
 4 years ago
Help Please :D

This Question is Closed

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0What happens if you multiply by P from the left?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0uhhh i thought P*P^1=Identity matrix

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0oh ya maybe I am wrong

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0No it equals Identity matrix

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[A^{1} P A = D\] \[AA^{1} P A = PA = A D\] \[P A A^{1} = P = A D A^{1} \]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Oh crap I used the wrong variables. Just substitute P for A ;)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0noooooooo it was P^1AP=D

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0For the second part \(A \neq D \)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0This is known as http;//en.wikipedia.org/wiki/Diagonalizable_matrix

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I so dont get what he did

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Here D is the diagonal matrix.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ya? how did u get that?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[ P^{1}AP = D \implies A = PDP^{1} \]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Using the right variables this time... \[P^{1} A P = D\] We can multiply by P from the left to get \[ P P^{1} A P = PD\] but \[P P^{1} = I\] so that means \[AP = PD\] Do the same thing on the right side with the inverse: \[AP P^{1} = PD P^{1}\] but since \[P P^{1} = I\] then \[A = P D P^{1} \]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0So how are we able to ignore the I?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0It's just the matrix equivalent of the number one.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Because \( I \) is the identity matrix.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0with regard to multiplication.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0^ Identity, not inverse :)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0LOL Thanks foolformath and jemurray

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Oh yes Identity not inverse.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0So from what i see A=d

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0It could be, but it doesn't have to be.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0D is a diagonal matrix, A may be not.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0huh? not getting it lol

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0What makes you think A = D?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0cuz I plugged this into the original equation

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1327223222985:dw

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0LOL I dont think that made sense so i take back what i said

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0There is no evidence that those two matrices are equal.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0That was really kinf o fyou to sit on my problem that long

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I am serious. I know it is late and .......

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Thanks i really appreciate it
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