anonymous
  • anonymous
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
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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amistre64
  • amistre64
greenday will have to help you since they feel noone is capable of helping out ...
amistre64
  • amistre64
D is just the eugene values on the identity
amistre64
  • amistre64
I cant recall for the life of me what to do with the P tho

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amistre64
  • amistre64
do you recall how to find the eugene values?
phi
  • phi
It is asking for the eigenvalues (diagonal entries of D) and eigenvectors (cols of P)
amistre64
  • amistre64
−1-L 4 −2 −3 4-L 0 −3 1 3-L determinate of that and solve for L
amistre64
  • amistre64
i spose we could echlon it too to find out but i never get a good result that way
amistre64
  • amistre64
eugene vectors; thats the P .... ill remember that someday lol
amistre64
  • amistre64
−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?
amistre64
  • amistre64
−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
amistre64
  • amistre64
L=-1 1+6+11+6 not= 0 ugh!!
amistre64
  • amistre64
maybe the eugenes are the negative diags? or i simply mismathed the whole thing
JamesJ
  • JamesJ
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.
amistre64
  • amistre64
if we echoln the matrix, dont we get the Evalues as well? or is that just me wishful thinking?
JamesJ
  • JamesJ
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.
amistre64
  • amistre64
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 :)
amistre64
  • amistre64
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
JamesJ
  • JamesJ
that's a big help.
amistre64
  • amistre64
i know L=1 works by sheer luck, so whats left over is a quadratic
anonymous
  • anonymous
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
amistre64
  • amistre64
yeah, ill have to go over finding Evectors again to refresh me cobwebs :)
amistre64
  • amistre64
it may be from the augmented | A LI| gauss jordon stuff
amistre64
  • amistre64
(A+LI)X = 0
phi
  • phi
Once you find an eigenvalue, subtract it from the diagonal, and find the null space of the resulting matrix. Use elimination to do this
amistre64
  • amistre64
or simply −1+L 4 −2 −3 4+L 0 −3 1 3+L row reduced for each L
phi
  • phi
to find the evals solve −1-L 4 −2 det −3 4-L 0 = 0 −3 1 3-L as previously posted
amistre64
  • amistre64
http://www.sosmath.com/matrix/eigen2/eigen2.html might be the same, hard for me to tell
phi
  • phi
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
amistre64
  • amistre64
i see, AX = LX AX - LX = 0 (A-LI)X = 0
amistre64
  • amistre64
they had a L = -4 which threw me for a loop :)

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