
Diogenes F. answered 05/02/21
Advanced Physics and Mathematics Tutor
Since you need to find the transformed quadratic form it suffices to find the eigenvalues of the quadratic form in x coordinates:
Q = (1,-1,-1,-1),(-1,3,0,0),(-1,0,7,-4),(-1,0,-4,7))
The characteristic polynomial can be found by evaluating the determinant |Q - λI| and its roots are the eigenvalues. In this case the eigenvalues are:
E = (11,4,3,0)
While the eigenvectors are:
V11 = (0,0,-1,1)
V4 = (-1,1,1,1)
V3 = (0,-2,1,1)
V0 = (3,1,1,1)
The matirx P is then P = (V11,V4,V3,V0) and in this variables the quadratic form is:
Qy = ((11,0,0,0),(0,4,0,0),(0,0,3,0),(0,0,0,0))
yt Q y = 11 y12 + 4 y22 + 3 y32