I want to show that if A(nxn) is symmetric then it is orthogonally equivalent to a symmetric matrix wit equal elements in the diagonal.For example for 2x2 if A=[a b;b c] there exist a Q O(2) (Q is orthogonal matrix)such thet :
Q 'AQ=[d e;e d]
Something which I have found till now :
1-In case 2x2 if orthogonal matrix is Q= [cos(teta) -sin(teta);sin(teta) cos(teta)] and A=[a b;b c] then for teta=.5*atan(c-a)/2b Q'AQ=[ f h;h f].
2-Since A is symmetric then it can be written as A=V'YV such that Y=V'AV and Y is a diagonal mat with eigen value on its diagonal and then same story for Y which is easier to find a Q matrix and then define a function f same as above and ....
3-It is possible that we may not find a general form for nxn.We may define f:R(nxn)----->R f(x)=SUM(yii-yjj) , Y=X'AX and note that f is a continuous function.And then we have to prove that f has a global minima Q* in O(n) and for that f(Q*)=0