## KonradZuse 3 years ago diagonalizing a matrix

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?

3. UnkleRhaukus

$S=P^{-1}$?

crap I forgot to post what the book said h/o.

@UnkleRhaukus

7. UnkleRhaukus

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...

10. UnkleRhaukus

mhm yeah I entered that in wolfram, but I'm not sure what SJ and M are... They are different from P and A....

@lgbasallote

13. anonymous

14. UnkleRhaukus

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.

:(

21. 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

22. 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.