Question

what is diagnolization in linear algebra?

Answer 1

I'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

Answer 2

thanks

Answer 3

@Callisto @Chlorophyll how are you guys with linear algebra? wanna help?

Answer 4

im totally lost. fml. my final is tomoro!!

Answer 5

Why don't you ask a specific question? Your question is too general and it takes time to answer all aspects of it.

Answer 6

okay: 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]

Answer 7

i dont even understand what to do here!

Answer 8

First find
\[Det( A - \lambda I)

\]

Answer 9

of what? A?

Answer 10

A is your matrix.

Answer 11

bec the 3 matrix udner it is: PD(P-1)

Answer 12

\[
Det(A-\lambda I)=-\lambda ^3+8 \lambda ^2-21
\lambda +18=(\lambda -3)^2 (2-\lambda )
\]
So it has 3 as a double eigenvalue and 2.

Answer 13

\[-\lambda^3+8\lambda^2-20\lambda+18\]

Answer 14

-21 instead of -20

Answer 15

dang it. how? isnt it 4 lambda 8 and 8 ?

Answer 16

Now we have to find the eigenspace of 2 and the eigenspace of 3

Answer 17

okay so how do we do it?

Answer 18

A-3? A-2I?

Answer 19

Find 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

Answer 20

wooahhh. hold on. how did u do that? where are those y1 vextor from??

Answer 21

I have to go to sleep now. Try to unersatnd how I got what I got. Bye.

Answer 22

thnaks