Question

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 )

Answer 1

i wrote in my notitions , you bring your quadratic form back to your diagonalised form but how do you do that ?

Answer 2

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

Answer 3

yes it's a symmetric matrix so inverse is the transposed. that figured it out !! thank you :)