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anonymous
 3 years ago
Let A = [3,2,6
0,1,6
0,0,3]
Find an invertible matrix P and a diagonal matrix D such that D = P^{1}AP.
find P and D
I worked out the polynomial 7x^37x^215x9 to give x= 1,3,3
My answer for P= [1,0,0
1,3,3
0,1,1]
and D= 1,0,0
0,3,0
0,0,3
something's wrong!
anonymous
 3 years ago
Let A = [3,2,6 0,1,6 0,0,3] Find an invertible matrix P and a diagonal matrix D such that D = P^{1}AP. find P and D I worked out the polynomial 7x^37x^215x9 to give x= 1,3,3 My answer for P= [1,0,0 1,3,3 0,1,1] and D= 1,0,0 0,3,0 0,0,3 something's wrong!

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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0what type of decomposition are you doing here ?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I think you get something wrong at P, since eigenvalue 3 has 2 dimensions and they are not yours! recheck

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0you should get D=dia(3,1,3)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0yes, surely, because they are eigenvalues of A, what I need is checking the order this guy arrange in P to have corresponding of numbers, like (3,1,3) or (1,3,3)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0We cannot guess, right, kid? if we mess the order up, although we know it is the Diagonal matrix of A but the P is in different order of eigenvectors, We will be crazy when checking what wrong is, I had that experience. It made me crazy

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0tr(A)=313=7 A11+A22+A33=3+9+3=15 A=3(3)=9 so, the characteristic polynomial would be\[\Delta(t)=t^3+7t^2+15t+9\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0well, as long as the eigen vector columns are arranged in proper order, you should be good.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0eigen vectors=? for t=1: \[2x+2y+6z=0\\6z=0\\3z=0\implies v_1=(1,1,0)^T\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0and v2 and v3 I put at (0,3,1)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0for t= 3: \[ 2y+6z=0\\ 2y+6z=0\\ v_{2,3}=(\alpha,3\beta,\beta)^T \]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0@thdrbird you cannot have them two identical

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0you can use v2 when a=0,b=1 v3 when a=1,b=1

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0or v3 when a=1,b=0 ANY combination of alpha and beta that do not give the same vectors..

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0can you give me an example?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I forgot to set alpha!

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0thanks @electrokid, great help!

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0@Hoa as well! Thanks!
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