At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
greenday will have to help you since they feel noone is capable of helping out ...
D is just the eugene values on the identity
I cant recall for the life of me what to do with the P tho
do you recall how to find the eugene values?
It is asking for the eigenvalues (diagonal entries of D) and eigenvectors (cols of P)
−1-L 4 −2 −3 4-L 0 −3 1 3-L determinate of that and solve for L
i spose we could echlon it too to find out but i never get a good result that way
eugene vectors; thats the P .... ill remember that someday lol
−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?
−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
L=-1 1+6+11+6 not= 0 ugh!!
maybe the eugenes are the negative diags? or i simply mismathed the whole thing
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.
if we echoln the matrix, dont we get the Evalues as well? or is that just me wishful thinking?
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.
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 :)
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
that's a big help.
i know L=1 works by sheer luck, so whats left over is a quadratic
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
yeah, ill have to go over finding Evectors again to refresh me cobwebs :)
it may be from the augmented | A LI| gauss jordon stuff
(A+LI)X = 0
Once you find an eigenvalue, subtract it from the diagonal, and find the null space of the resulting matrix. Use elimination to do this
or simply −1+L 4 −2 −3 4+L 0 −3 1 3+L row reduced for each L
to find the evals solve −1-L 4 −2 det −3 4-L 0 = 0 −3 1 3-L as previously posted
http://www.sosmath.com/matrix/eigen2/eigen2.html might be the same, hard for me to tell
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
i see, AX = LX AX - LX = 0 (A-LI)X = 0
they had a L = -4 which threw me for a loop :)