Similarity Transformations

(2x2 real matrix)

  Matrix A   Matrix S  
  (a) (b)   (u) (v)  
  (c) (d)   (w) (y)  
  Δ(det): tr:   Δ(det): tr:  
  (tr/2)2   (tr/2)2  
  anti-tr: anti-tr/2b   anti-tr: anti-tr/2v  
  √[(tr/2)2 -Δ] +i   √[(tr/2)2 -Δ] +i  
  λ1 +i   λ1 +i  
  λ2 +i   λ2 +i  
  (y/x)1: +i   (y/x)1: +i  
  (y/x)2: +i   (y/x)2: +i  
  (y/x): +i   (y/x): +i  
  Matrix K=SAST     Matrix Ka=STAS    
  (K11) (K12)   (Ka11) (Ka12)  
  (K21) (K22)   (Ka21) (Ka22)  
  Δ(det): tr:   Δ(det): tr:  
  (tr/2)2   (tr/2)2  
  anti-tr: anti-tr/2k12   anti-tr: anti-tr/2ka12  
  √[(tr/2)2 -Δ] +i   √[(tr/2)2 -Δ] +i  
  λ1 +i   λ1 +i  
  λ2 +i   λ2 +i  
  (y/x)1: +i   (y/x)1: +i  
  (y/x)2: +i   (y/x)2: +i  
  (y/x): +i   (y/x): +i  
  Find Similarity matrix satisfying condition (2) & (4)      
  (a1) (b1)   (u1) (v1)  
  (c1) (d1)   (w1) (y1)  
  Find Similarity matrix satisfying condition (1) & (3)   b=c ; d/a = u/y  
        (1-ad/b2)=± i    
  (a1a) (b1a)   (u1a) (v1ar) - i(v1ai)  
  (c1a) (d1a)   (w1ar) + i(w1ai) (y1a)  
  Matrix L=SAS-1     Matrix La=S-1AS    
  (L11) (L12)   (La11) (La12)  
  (L21) (L22)   (La21) (La22)  
  (a2) (b2)   +i (u2) +i (v2)  
        + i(u2) + i(v2)  
  (c2) (d2)   +i (w2) (y2)  
        + i(w2)    


* Ka=STAS = u  w   *   a   b   * u   v    = u2a+w2d+uw(b+c)          (auv+dwy)+cvw+buy

                       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

                  0               w2a+y2d+2wyb

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.

Since K=Ka,

Ka=  u2a+w2d                           (auv+dwy)+cvw+buy

         (auv+dwy)+bvw+cuy       v2a+y2

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

b=-c .....(5)

v=w ....(6)

w/y =u/v=  √(-d/a) =k


u=vk .....(7)

w=yk  ....(8)

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

b=c ........(9)


w+v=-2bu/d ....(11)


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.