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anonymous
 one year ago
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 )
anonymous
 one year ago
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 )

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i wrote in my notitions , you bring your quadratic form back to your diagonalised form but how do you do that ?

phi
 one year ago
Best ResponseYou've already chosen the best response.1as 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
 one year ago
Best ResponseYou've already chosen the best response.0yes it's a symmetric matrix so inverse is the transposed. that figured it out !! thank you :)
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