(2x2 real matrix)
|* Ka=STAS = u w * a
b * u v = u2a+w2d+uw(b+c)
v y c d w y (auv+dwy)+bvw+cuy v2a+y2d+vy(b+c)
* K=SAST = u v * a b * u w = u2a+v2d+uv(b+c) (auw+dvy)+cvw+buy
w y c d v y (auw+dvy)+bvw+cuy w2a+y2d+wy(b+c)
We have to find such S so that K=Ka= a diagonal matrix. Such transformation is a similarity transformation . Let us find out the nature of S.
To be diagonal , wvb+cuy = wvc+buy or wv(b-c) = uy(b-c)
or b-c=0 => b=c ....(1)
or wv=uy ....(2)
and uwa+yvd=-(wvb+yuc)=-b(wv+yu) if b=c ...(3)
or uwa+yvd=-yu(b+c) if wv=uy ....(4)
or uwa+yvd=-2yub if both (1) and (2) are satisfied.
If condition (1) & (3) is satisfied,
K= u2a+v2d+2uvb 0
If condition (2) & (4) is satisfied,
K= u2a+v2d+uv(b+c) 0 if b+c=0 , then K = u2a+v2d 0
0 w2a+y2d+wy(b+c) 0 w2a+y2d
How to simplify condition (2) and (4) :-
For K12=0 , (uwa+yvd) +(cvw+buy) =(uwa+yvd) + b(yu-wv)=0
For K21=0, (uwa+yvd) - b(yu-wv)=0
which means (uwa+yvd)=0 and yu-wv=0, hence u/v =w/y
(uwa+yvd)=0 implies that v/y(w2a+y2d)=0 or w/y = √(-d/a)
So the conditions are : b=-c , u/v=w/y = √(-d/a) and d,a should be of opposite sign for real v,w.
Ka= u2a+w2d (auv+dwy)+cvw+buy
Ka12=(auv+dwy) + b(yu -wv) =0
Ka21=(auv+dwy) - b(yu -wv) =0 , (yu -wv) =0 means u/v =w/y
auv+dwy=0 means w/y(v2a+y2d ) =0 or v/y= √(-d/a) . Since as proved above, w/y= √(-d/a), so v=w
Hence overall conditions are
w/y =u/v= √(-d/a) =k
First take a value of y. Then give a value to d and a(of opposite sign) and find k. Then Find w. Then find v. Then find u. The product matrix of such similarity transformation is a null matrix.
Arbitrary choice of a,d,b,y
When Condition (1) and (3) are satisfied :-
we assume b=c
K = u2a+v2d+2buv (auw+dvy)+b(vw+uy) Ka= u2a+w2d+2buw (auv+dwy)+b(vw+uy)
(auw+dvy)+b(vw+uy) w2a+y2d+2bwy (auv+dwy)+b(vw+uy) v2a+y2d+2bvy
Now (comparing corresponding elements of K and Ka,=
auw+dvy=auv+dwy or au(w-v) = dy(w-v) or au=dy
now u2a+w2d+2buw =u2a+v2d+2buv or d(w2-v2)=2bu(v-w) or d(w+v) =-2bu or w+v=-2bu/d
Now (auw+dvy)+b(vw+uy) =0 or (dyw+dvy) +b(vw+uy)=0 or dy(v+w) +b(vw+uy)=0 or -2buy +buy+bvw =0 or bvw=buy or vw=uy=uua/d =u2a/d
v-w = (v+w)2 -4vw =4b2u2/d2 -4u2a/d = (2u/d) √( b2-ad)
So v=(-bu/d)[1 -√ (1 - ad/b2)]
w=(-bu/d)[1 +√ (1 - ad/b2)]
let m =(1 - ad/b2)
When Δ(det)=ad/b2 < 0 , m >1, v=(-bu/d)[1 -√ m], figure under blue is -ve. w=(-bu/d)[1 +√ m], figure under blue is +ve.So v,w are of opposite sign.
When Δ(det)=ad/b2 =0 , m=1, v=0, w=(-2bu/d)
when 0< Δ(det)=ad/b2 < 1, 0 <m < 1, v=(-bu/d)[1 -√ m],w=(-bu/d)[1 +√ m] . So v,w are of same sign.
when Δ(det)=ad/b2=1, m=0, v=w=-bu/d
when Δ(det)=ad/b2 > 1, m is imaginary no. say i√|m| , v =(-bu/d)[1 -i√ |m|]
w=(-bu/d)[1 +i√ |m|] i.e. v, w are complex conjugates of each other.
Summary of condition (1) , (3) being satisfied in special case when b=c
First take a value of y. Take a value of b,c such that b=c.Take a value of d,a. Then find value of u. Then find w,v.
Arbitraty choice of a,d,b,y
|* If A is a symmetric matrix, and S is any invertible matrix
satisfying K(diagonal)=SAST, then
(1) no. of negative elements in K is always the same for all such S , and same goes for positive elements.
(2) A, K are congruent.
(3) If A is co-efficient matrix of some quadratic form of Rn , then K is the matrix for the same form after change of basis defined by S.
(4) A symmetric matrix A can always be transformed in this way into a diagonal matrix having only entries 0,+1,-1 .
(5) The no. of diagonal entries of each sign is an invariant of A, it does not depend on the matrix S used.
(6) No. of 1's : denoted by n+ -> positive index of inertia
No. of -1's : denoted by n- ->negative index of inertia
No. of 0's : denoted by n0 -> dimension of null space
n0+n+ +n- =n
n+ -n- = signature of A =Sign(A)
(7) The law is known as Sylvester's Law of Inertia and can be restated as : Two symmetric matrices of the same size have the same number of +, - , zero eigen values iff they are congruent.
(8) If AAT = ATA or A+A =AA+, A is called a Normal matrix.