Question

why is S^T A S = Diagonal matrix
if A is n by n symmetric and
1)There exists n orthonormal eigen vectors{e1,e2,e3...en}, such that (ei,ej)=sigma(i,j)
2)There exists an orthogonal transformation matrix S =[e1,e2,e3..en] such that the det(S)=+/- 1 and S^t=s^-1
3)There exist an orthogonal similarity transformation: For all X E R^n x=y1e1+y2e2+...+ynen =Sy
Such that s^TAS=D
D is a diagonal matrix of eigen values

Answer 1

@ganeshie8

Answer 2

trying to solve this problem 2x^2-4xz+z^2-4yz=4
wanna turn this into another linear combinatons of squares only

Answer 3

by rewriting x,y,z, in principical axis

Answer 4

because det(S) = +/- u can say this is a principical basis

Answer 5

+/-1

Answer 6

*

Answer 7

this is all relevant because first of all
where vector x=[x1,x2,x3] Q(x)=(x,Ax)=x^TAx
so if we can rewrite
Q(x)=(x,Ax)=Q(Sy)=(Sy,ASy)=S^Ty^T ASy =(y,S^TASy)
not S^TAS = D they say so, why? but if it is so we can see how we know have a pure lnear combination of squares
as
(y,Dy)=j from 1 to n Sum( Lambdaj Yj^2

Answer 8

instead of using x,,y,z i made it
x=x1
y=x2
z=x3

Answer 9

\(\large S^{T} = S^{-1}\) if the eigen vectors chosen are orthogonal

Answer 10

i cant believe we are back to this alll over lol!! Why are these eigen vectors orthogonal

Answer 11

lol i remember you asked that question earlier and i tried to make sense of it myself geometrically but its hard to visualize

Answer 12

I have a theorem that states Eigen Vectors for distinct eigen values are orthogonal and Eigen vectors for multiple eigen values can be orthogonalized wit hthe gramschimit process

Answer 13

so i had to accept the analytic proof to move on

Answer 14

yeah u are right i shud just finish this assignment and then mess around trying to prove this stuff lol