• anonymous
I'm studying Quadratic forms now and i have this theorem that i don't understand fully it says : for a quadratic form Q(x)= X'AX, there is an orthogonal matrix P wherefore Y=P'X: X'AX=Y'DY=a1²y1²+...+ an²yn² can anybody explain Y=P'X ? ( the accent is the transposed matrix mostly in english textbooks it is given by a T )
  • Stacey Warren - Expert
Hey! We 've verified this expert answer for you, click below to unlock the details :)
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.
  • chestercat
I got my questions answered at in under 10 minutes. Go to now for free help!
  • anonymous
i wrote in my notitions , you bring your quadratic form back to your diagonalised form but how do you do that ?
  • phi
as near as I can tell, they diagonalize matrix A if we have n independent eigenvectors of A and put them in S then we can write the eigenequations in the form \[ A S = \Lambda S \\ A S = S\Lambda \\ A= S\Lambda S^{-1} \] where \( \Lambda \) is the diagonal matrix of eigenvalues if we use that in place of A we have \[ X^T A X = X^T S\Lambda S^{-1} X \] it looks like \( Y= S^{-1} X \) though I may be missing something, as that shows an inverse rather than a transpose of the eigenvector matrix
  • anonymous
yes it's a symmetric matrix so inverse is the transposed. that figured it out !! thank you :)

Looking for something else?

Not the answer you are looking for? Search for more explanations.