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thdrbird
 one year 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!
thdrbird
 one year 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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electrokid
 one year ago
Best ResponseYou've already chosen the best response.2what type of decomposition are you doing here ?

Hoa
 one year ago
Best ResponseYou've already chosen the best response.1I think you get something wrong at P, since eigenvalue 3 has 2 dimensions and they are not yours! recheck

electrokid
 one year ago
Best ResponseYou've already chosen the best response.2you should get D=dia(3,1,3)

Hoa
 one year ago
Best ResponseYou've already chosen the best response.1yes, 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)

Hoa
 one year ago
Best ResponseYou've already chosen the best response.1We 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

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

electrokid
 one year ago
Best ResponseYou've already chosen the best response.2well, as long as the eigen vector columns are arranged in proper order, you should be good.

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

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

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

electrokid
 one year ago
Best ResponseYou've already chosen the best response.2@thdrbird you cannot have them two identical

electrokid
 one year ago
Best ResponseYou've already chosen the best response.2you can use v2 when a=0,b=1 v3 when a=1,b=1

electrokid
 one year ago
Best ResponseYou've already chosen the best response.2or v3 when a=1,b=0 ANY combination of alpha and beta that do not give the same vectors..

thdrbird
 one year ago
Best ResponseYou've already chosen the best response.0can you give me an example?

thdrbird
 one year ago
Best ResponseYou've already chosen the best response.0I forgot to set alpha!

thdrbird
 one year ago
Best ResponseYou've already chosen the best response.0thanks @electrokid, great help!

thdrbird
 one year ago
Best ResponseYou've already chosen the best response.0@Hoa as well! Thanks!
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