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diagonalizing a matrix

Linear Algebra
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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...
@UnkleRhaukus any idea sir?
\[S=P^{-1}\]?

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Other answers:

crap I forgot to post what the book said h/o.
1 Attachment
whats your question?
I'm trying to do #14.
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...
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
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 not my habit to reply to questions already being answered
i think that S and P are inverses of each other
that's what I was looking at. @iggy we are in the process yo...
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....?
err # 13.
yeah Idk about S and P...
it look slike J = P^-1AP just flipped around... Same thing happened when I did #15.
:(
  • 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
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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