Question

hey im struggling with a qustion, can someone help please?

if A = the matrices:
−1 4 −2
−3 4 0
−3 1 3

Find an invertable matrix P and diagonal matrix D so that AP = PD.

thank you for any feedback x

Answer 1

greenday will have to help you since they feel noone is capable of helping out ...

Answer 2

D is just the eugene values on the identity

Answer 3

I cant recall for the life of me what to do with the P tho

Answer 4

do you recall how to find the eugene values?

Answer 5

It is asking for the eigenvalues (diagonal entries of D) and eigenvectors (cols of P)

Answer 6

−1-L 4 −2
−3 4-L 0
−3 1 3-L

determinate of that and solve for L

Answer 7

i spose we could echlon it too to find out but i never get a good result that way

Answer 8

eugene vectors; thats the P .... ill remember that someday lol

Answer 9

−1 4 −2
−3 4 0
−3 1 3


−1 4 −2
0 -8 6
0 -11 9


−1 4 −2
0 -8 6
0 0 6/8

maybe?

Answer 10

−1-L 4 −2
−3 4-L 0
−3 1 3-L

(-1-L)( (4-L)(3-L)-0)
+3 (4(3-L)-1()-2)
-3(0-(4-L)(-2)


(-1-L)(L^2-7L+12)
36-12L+6
-24+6L


-L^3+6L^2-11L+6

with any luck thats the eugene equations

Answer 11

L=-1
1+6+11+6 not= 0 ugh!!

Answer 12

maybe the eugenes are the negative diags? or i simply mismathed the whole thing

Answer 13

Yes, the matrix D has the eigenvalues of A down it's diagonal. So first you need to find the eigenvalues.

Then the matrix P is the matrix of the eigenvectors as column vectors corresponding to those eigenvalues.

Answer 14

if we echoln the matrix, dont we get the Evalues as well? or is that just me wishful thinking?

Answer 15

No, we don't. There's no escaping solving the characteristic equation.

@sammy_maths: I'm sure you have a worked example of a problem like this in your class notes, albeit for a 2x2 or 3x3 matrix. I can't find a good one to link to right now. But go back and have a look at it. And then if you're still stuck, as us again.

Answer 16

bummer...

http://www.wolframalpha.com/input/?i=eigenvalues+%7B%7B-1%2C4%2C-2%7D%2C%7B-3%2C4%2C0%7D%2C%7B-3%2C1%2C3%7D%7D

you can use this for a check :)

Answer 17

http://www.wolframalpha.com/input/?i=characteristic+polynomial%7B%7B-1%2C4%2C-2%7D%2C%7B-3%2C4%2C0%7D%2C%7B-3%2C1%2C3%7D%7D

yay, my char eq was right lol

Answer 18

that's a big help.

Answer 19

i know L=1 works by sheer luck, so whats left over is a quadratic

Answer 20

I have been following it, and I'm understanding where you get the values from so far
The only notes that may help now is
the answer to P =
1 2 1
3 3 1
4 3 1

thank you x

Answer 21

yeah, ill have to go over finding Evectors again to refresh me cobwebs :)

Answer 22

it may be from the augmented | A LI| gauss jordon stuff

Answer 23

(A+LI)X = 0

Answer 24

Once you find an eigenvalue, subtract it from the diagonal, and find the null space of the resulting matrix. Use elimination to do this

Answer 25

or simply

−1+L 4 −2
−3 4+L 0
−3 1 3+L

row reduced for each L

Answer 26

to find the evals solve
−1-L 4 −2
det −3 4-L 0 = 0
−3 1 3-L
as previously posted

Answer 27

http://www.sosmath.com/matrix/eigen2/eigen2.html

might be the same, hard for me to tell

Answer 28

it is easy with matlab or wolfram, and tedious by hand
(-1-L)(4-L)(3-L) - 4(-3(3-L)) -2(-3 +3(4-L)) = 0

Answer 29

i see,

AX = LX

AX - LX = 0

(A-LI)X = 0

Answer 30

they had a L = -4 which threw me for a loop :)