KonradZuse
  • KonradZuse
diagonalizing a matrix
Linear Algebra
  • Stacey Warren - Expert brainly.com
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
KonradZuse
  • KonradZuse
Trying to figure out how this works, wolfram is giving me a different setup than my book... http://www.wolframalpha.com/input/?i=diagonalize+%7B%7B1%2C0%7D%2C%7B6%2C-1%7D%7D The book says p = [1/3][0] [1 ][1] and P^-1AP = [1][0] [0][-1] Which is the inverse of J...
KonradZuse
  • KonradZuse
@UnkleRhaukus any idea sir?
UnkleRhaukus
  • UnkleRhaukus
\[S=P^{-1}\]?

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KonradZuse
  • KonradZuse
crap I forgot to post what the book said h/o.
KonradZuse
  • KonradZuse
1 Attachment
KonradZuse
  • KonradZuse
@UnkleRhaukus
UnkleRhaukus
  • UnkleRhaukus
whats your question?
KonradZuse
  • KonradZuse
I'm trying to do #14.
KonradZuse
  • KonradZuse
I'm confused on how to do this... Wolfram gives a different definition than my book, and I'm not really sure what my book is trying to show...
UnkleRhaukus
  • UnkleRhaukus
http://www.wolframalpha.com/input/?i=diagonalise+matrix+%7B%7B1%2C0%2C0%7D%2C%7B+0%2C1%2C1%7D%2C%7B0%2C1%2C1%7D%7D
KonradZuse
  • KonradZuse
mhm yeah I entered that in wolfram, but I'm not sure what SJ and M are... They are different from P and A....
KonradZuse
  • KonradZuse
@lgbasallote
lgbasallote
  • lgbasallote
@KonradZuse not my habit to reply to questions already being answered
UnkleRhaukus
  • UnkleRhaukus
i think that S and P are inverses of each other
KonradZuse
  • KonradZuse
that's what I was looking at. @iggy we are in the process yo...
KonradZuse
  • KonradZuse
but according to wolfram and the example 7 I showed above, there is no 1/3 in wolfram.... Is it just a different way to do it....?
KonradZuse
  • KonradZuse
err # 13.
KonradZuse
  • KonradZuse
yeah Idk about S and P...
KonradZuse
  • KonradZuse
it look slike J = P^-1AP just flipped around... Same thing happened when I did #15.
KonradZuse
  • KonradZuse
:(
phi
  • phi
diagonalizing a matrix M uses eigenvectors x and eigenvalues \(\lambda\) the "big equation" is \( M x = \lambda x\) if we put all the eigenvectors x into the columns of a matrix P and the eigenvalues on the diagonal of matrix \( \Lambda \) we can say \( MP = P\Lambda \) note the order we must multiply P and \( \Lambda \) to get the right result multiplying by the inverse of P we have \(\Lambda = P^{-1}MP \) This says we can diagonalize matrix M by multiplying by the eigenvector matrix P It returns a diagonal matrix with entries being the eigenvalues of M
phi
  • phi
When we do this diagonalizing, note that there is flexibility in defining an eigenvector. Eigenvectors represent a "direction", and if we scale them, they are still an eigenvector. Example: if eigenvector x= [ 1/3 1] corresponds to eigenvalue λ and we scale x by 3 to get x' = [1 3] , it is still an eigenvector associated with λ M(3x)=λ(3x) still works. We would scale x to get rid of fractions. Also, when diagonalizing, there is no forced order on the eigenvalues in matrix \(\Lambda\) The only requirement is that the eigenvalue in position n,n of matrix \(\Lambda\) corresponds to the eigenvector in column n of matrix P Often, though, people put the eigenvalues in ascending (or descending ) order. The point is we can get different variations of P and Λ, and all are correct, unless the order of the eigenvalues in Λ are stipulated to be in some order.

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