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TuringTest
 2 years ago
Best ResponseYou've already chosen the best response.1I'm kind of weak in this area, so I'm just going to shoot you a link http://tutorial.math.lamar.edu/Classes/LinAlg/Diagonalization.aspx I'm sure somebody can explain it in their own words though... @UnkleRhaukus @nbouscal

TuringTest
 2 years ago
Best ResponseYou've already chosen the best response.1@Callisto @Chlorophyll how are you guys with linear algebra? wanna help?

liliy
 2 years ago
Best ResponseYou've already chosen the best response.0im totally lost. fml. my final is tomoro!!

eliassaab
 2 years ago
Best ResponseYou've already chosen the best response.1Why don't you ask a specific question? Your question is too general and it takes time to answer all aspects of it.

liliy
 2 years ago
Best ResponseYou've already chosen the best response.0okay: use the diagonlization theomr to find eignevalues of A and baisis for each eignespace: A= [ 2 1 1 1 4 1 1 1 2] =[ 1 1 0 [ 3 0 0 [ 0 1 1 1 1 1 0 2 0 1 1 1 0 1 1] 0 0 3] 1 1 0]

liliy
 2 years ago
Best ResponseYou've already chosen the best response.0i dont even understand what to do here!

eliassaab
 2 years ago
Best ResponseYou've already chosen the best response.1First find \[Det( A  \lambda I) \]

liliy
 2 years ago
Best ResponseYou've already chosen the best response.0bec the 3 matrix udner it is: PD(P1)

eliassaab
 2 years ago
Best ResponseYou've already chosen the best response.1\[ Det(A\lambda I)=\lambda ^3+8 \lambda ^221 \lambda +18=(\lambda 3)^2 (2\lambda ) \] So it has 3 as a double eigenvalue and 2.

liliy
 2 years ago
Best ResponseYou've already chosen the best response.0\[\lambda^3+8\lambda^220\lambda+18\]

liliy
 2 years ago
Best ResponseYou've already chosen the best response.0dang it. how? isnt it 4 lambda 8 and 8 ?

eliassaab
 2 years ago
Best ResponseYou've already chosen the best response.1Now we have to find the eigenspace of 2 and the eigenspace of 3

eliassaab
 2 years ago
Best ResponseYou've already chosen the best response.1Find X such that AX = 2 X and Y such AY =3Y Y1={1, 0, 1} Y2={1, 1, 0} X={1, 1, 1} Y1, and Y2 is a basis for the eigenspace of 3 X is a basis for the eigensapce 2

liliy
 2 years ago
Best ResponseYou've already chosen the best response.0wooahhh. hold on. how did u do that? where are those y1 vextor from??

eliassaab
 2 years ago
Best ResponseYou've already chosen the best response.1I have to go to sleep now. Try to unersatnd how I got what I got. Bye.
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