* Ka=S^TAS = 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=SAS^T = 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 is  a diagonal matrix. Such transformation is a similarity transformation . Let us find out the nature of S.

K to be diagonal ,

(auw+dvy) +cvw+buy= 0 .............(1)

(auw+dvy) +cvw+buy=(auw+dvy)+bvw+cuy  or uy=vw......(2) This is equality condition of both diagonal elements

From (1) auw+dvy+uy(b+c)=0 ; Substituting the value of v=uy/w , we get

aw² +dy²+wy(b+c)=0 for condition (1) to be satisfied. this also means that K22 is zero.

One can arbitrarily fix the value of either w or y . In either case, this is a quadratic equation. Here we arbitrarily assign a value to y.

w=   (-y(b+c) ± √[y²(b+c)² - 4ady² ] )/2a

Arbitrarily assign a value to u. Then v=uy/w;

We have also  to find such S so that Ka is  a diagonal matrix. Such transformation is a similarity transformation . Let us find out the nature of S.

Ka to be diagonal ,

(auv+dwy) +cvw+buy= 0 .............(3)

(auv+dwy) +cvw+buy=(auv+dwy)+bvw+cuy  or uy(b-c)=vw(b-c)......(4) This is equality condition of both diagonal elements

so that uy=vw

From (3) auv+dwy+uy(b+c)=0 ; Substituting the value of w=uy/v , we get

av² +dy²+vy(b+c)=0 for condition (1) to be satisfied. this also means that Ka22 is zero.

One can arbitrarily fix the value of either v or y . In either case, this is a quadratic equation. Here we arbitrarily assign a value to y.

v=   (-y(b+c) ± √[y²(b+c)² - 4ady² ] )/2a

Arbitrarily assign a value to u. Then w=uy/v;

K=          0                0   if uy=vw, y,u is arbitrary, w is found from quadratic equn. v is then found out.                     

               0                0

when K is null, Ka is not necessarily Null.

Ka=          0                0   if uy=vw, y,u is arbitrary, v is found from quadratic equn. w is then found out.                     

                 0                0

When Ka is null,K is not necessarily Null.

 It is interesting to see that y,u are kept arbitrary in both K,Ka and formula of w in K is identical to formula of v in Ka.

 which implies that v,w can be interchanged without affecting invariance. When attempt is made to make K or Ka diagonal , both become Null matrices.

Out of 2 conditions in each of K,Ka, uy=wv is the common condition , other condition is interchangeable for (v,w)

Conclusion: In usual course, We cannot diagonalize a 2x2 square matrix A  by S^TAS =Ka since Ka becomes a null matrix with condition uy=vw.But from Equn. (4) , we can try the other condition b=c which makes the off diagonal elements equal without adhering to uy=vw.

* If b=-c , uy=-vw , d/a= - v²/y², then Ka becomes a matrix whose diagonal elements are 0 and other diagonal elements have some value. When b=-c, in Ka, Ka22 = v²a+y²d+vy(b+c)= = v²a+y²d. If Ka=0, then   d/a= - v²/y²,                                      

Now We try the other condition of Equ.4 which is b=c, i.e. Matrix A becomes a symmetric matrix.

* Ka=S^TAS = u  w   *   a   b   * u   v    = u²a+w²d+2uwb          (auv+dwy)+bvw+buy

                       v  y        c   d       w  y       (auv+dwy)+bvw+buy       v²a+y²d+2vyb

* K=SAS^T = u  v   *   a   b   * u   w    = u²a+v²d+2uvb               (auw+dvy)+bvw+buy

                     w  y      c   d      v   y       (auw+dvy)+bvw+cuy            w²a+y²d+2wyb

(auv+dwy)+bvw+buy =(auw+dvy)+bvw+cuy or (auv+dwy) =(auw+dvy) or au(w-v) = dy(w-v).....(5) or au=dy which is equality condition.

Now (auw+dvy)+bvw+cuy =0 or auw+auv +b(uy+wv) =0 or au(v+w) +b[uau/d  +wv]=0

or abu² +ad(v+w)u+bdwv=0 which is a quadratic equn of u. Solving for u, we get

u= -d(v+w) / 2b  ± √(a²d²(v+w)²- 4ab²wvd ) / 2ab

or u =-d(v+w) /( 2b) [ 1   ±√ [1 - 4b²wv/(ad(v+w)²) ]

One can arbitrarily choose (w,v) -By making v=1,w=1, we get

u=-(d/b) ± (d/b)√[ 1- (b²/ad)]

y=au/d

Important Observation:

(1) if v and w are different, then K and Ka are diagonal, but their diagonal elements are different.

(2)If v=w , then K,Ka commute. But their diagonal elements are not equal to eigen value of A matrix.

Now We try another condition :

From equn (5),  au(w-v) = dy(w-v) . we make w=v

auw+dvy+bvw+cuy=0 or  auv+dvy+bv²+buy=0 or bv²+(au+dy)v+buy =0

or v=w=-(au+dy)/2b  ±   ((au+dy)/2b) * √[ 1-  (4uyb²)/(au+dy)²] = [-(au+dy)/2b]* ( 1 ±  √[ 1-  (4uyb²)/(au+dy)²])

or or v =-(au+dy) /( 2b) [ 1   ±√ [1   -   4b²uy / (au+dy)² ]

Here, unlike the previous case,K & Ka are not only diagonal but also   K=Ka whether y=u or not.

Again Another Condition :

AS=(S^T)^-1Ka or  a    b * u   v  =  y/Δ    -w/Δ    * λ1   0     or      au+bw    av+by   =  y λ1/Δ        - wλ2/Δ    

                            c    d    w   y     -v/Δ     u/Δ        0    λ2             cu+dw    cv+dy      -v λ1/Δ           u λ2/Δ

au+bw=  y λ1/Δ    ....(7)

av+by= - wλ2/Δ    ...(8)

cu+dw= -v λ1/Δ    ...(9)

cv+dy=   u λ2/Δ   ....(10)

This can be expressed in the matrix form as

a        -λ1/Δ    b         0   *   u   =   0

0           b       λ2/Δ    a        y         0

c           0       d        λ1/Δ    w        0

-λ2/Δ    d       0          c        v        0

cau+cbw = cyλ1/Δ...(11)

cau+daw=-vaλ1/Δ...(12)

(11)-(12) =-w(ad-bc) =(cy+av)λ1/Δ , since ad-bc=Δ

w=-(cy+av)λ1/Δ²

Similarly from (8) ,(10) we get

y=(cw+av)λ2/Δ²

w/y = - (cy+av)*λ1 /λ2 (cw+av) =-k (cy+av)/(cw+av)  where k= λ1 / λ2

or cw² +avw +(kcy² +kavy)=0 which is a quadratic form and also a quadratic equation of w.

or w=(-av ±√ [a²v² -4c(kcy² + kavy)])/2c

-w/u= (av+by) / (cv+dy) or u = -w*(cv+dy)/(av+by)

Arbitrarily choose y,v and we normally choose y=1,v=1 . Take b=c

Then find w,u from above equations.

Exa- Take matrix     14  12

                                   12    5

Diagonality of Ka does not necessarily ensure diagonality of K.

Easy way to find similarity Matrix for a symmetric matrix where b=c.

Build Matrix A. Press submit button. Find (y/x)1 and (y/x)2 which are the ratios of y to x component of the vector. In the S matrix first column, put

1                                                1

(y/x)1   and second column      (y/x)2 . This is the similar matrix.

For symmetric matrices, S^TAS = Ka which is diagonal matrix with diagonal values being eigen values.

But in normal cases       S^-1AS = Ka  which implies that S^T =S^-1  which is the case for orthogonal matrices.  

                                                                             c  d     w    y     w   y       0      λ₂         cu+dw   cv+dy      wλ₁        yλ₂

We get 4 equations out of which 2 are independent. We arbitrarily fix the value of u,v. then

w= (u/b)(λ₁-a)

y=  (v/b)(λ₂-a) For convenience, we put u=1,v=1. (other values can also be put)

It will be found that the column of the similarity matrix are the (y/x) ratios of the eigen vectors of matrix A.

However, we find that in general La is not equal to L , nor L is diagonal .

(b)

Similarity matrix for a square matrix A satisfies the equation SAS^-1=L where L is a diagonal matrix having elements λ₁ and λ₂ which are the eigen values of A. We can also write SA=LS..(6) or 

 u  v   * a    b  = λ₁    0   *  u     v   or  au+cv    dv+bu   =uλ₁         vλ₁

 w y      c    d      0     λ₂      w   y         cy+aw    bw+dy     wλ₂        yλ₂

We get 4 equations out of which 2 are independent.

au+cv=uλ₁

bu+dv=vλ₁

aw+cy=wλ₂

bw+dy=yλ₂

Then

v= (u/c)(λ₁-a)

w=  (y/b)(λ₂-d) For convenience, we put u=1,y=1. (other values can also be put)

[A, B] =        bg -cf                       bh-be-df+af   =         bg-cf                  b(h-e) -f(d-a)    =      m               b(h-e) -f(d-a)   =     m        X  =    m      X'         

               dg-ag-ch+ce                      -(bg-cf)          -c(h-e) +g(d-a)             -(bg-cf)            -c(h-e) +g(d-a)     -m                      Y       -m       Y'     -m

where X, Y are 2x2 matrices.

X=    b         f

      (d-a)   (h-e) 

Y=    -c         -g

       (d-a)   (h-e) 

and corresponding adjoints are given by   

X'  =  (h-e)           -f

        -(d-a)           b

Y' =   (h-e)          g

        -(d-a)         -c

Hence in a 2x2 commutator matrix , trace is always zero i.e. the matrix is traceless. This is because tr(AB) = tr(BA) and this is so irrespective of whether the 2 matrices commute or not. When the matrices commute, bg=cf or b/f = c/g .

From above, we can say that a traceless matrix can be expressed as a commutation product of 2 square matrices.

Case 1: When [A,B] =0, X,Y are both singular matrices and m=0

Case 2: When b=c=0   [A,B] = 0             -f(d-a)  and determinant = fg(d-a)²

                                                g(d-a)           0

Case 3: When f=g=0  [A,B] = 0             b(h-e)  and determinant = bc(h-e)²

                                              -c(h-e)           0

Case 4: When b=c=f=g=0, [A,B] =   0      0

                                                           0      0

Case 5: When d=a,             [A,B] =  bg-cf          b(h-e)   and   determinant = -(bg-cf)²  +bc(h-e)²

                                                         -c(h-e)       -(bg-cf)

Case 6: When e=h,             [A,B] =  bg-cf          -f(d-a)   and   determinant = -(bg-cf)²  +fg(d-a)²

                                                         -g(d-a)       -(bg-cf)

Case 7: when X=Y or (d-a)/(h-e) = (b+c)/(g+f)              [A, B]=  bg-cf            X       determinant =  -(bg-cf)²  - [b(h-e)-f(d-a)]²

                                                                                                         X            -(bg-cf)

Case 8: when X=Y=0, then bg-cf=0 and [A,B]=0

                      Y    y

We want to convert it to a form D1 =X   x                  

                                                          y   Y

then  a    b *  D  =  D1

        c    d

Solving we get   a  = (Yx - Xy) / (XY-xy)

                          d = (Xy - Yx) / (XY-xy)

                          b  = (X² - x²) / (XY-xy)

                          c  = (Y² - y²) / (XY-xy)

Suppose the matrix is  0    X

                                   Y    0

we can transform it to a block diagonal matrix by

 0       X/Y  * 0   X     =X      0

Y/X     0       Y    0       0      Y

or 

0        K  *   0      X =   X     0      where K = X / Y

K^-1    0       Y      0       0     Y

or   T     *    0     X   =X      0

                   Y     0      0      Y 

T is the transformation matrix given by

0       K

K^-1    0

K matrix is given by

K = k11   k12       =    [b(h-e) - f*d-a)] / [-c(h-e) +g(d-a)]                 (bg-cf) / [-c(h-e) +g(d-a)]     and   KY=X         

       k21   k22                                    0                                                                1

We know that if the representation of a group is fully reducible, the same can be represented as a block matrix which is the direct sum of its irreducible sub-representations,

In this case, we have hit upon a block diagonal matrix. We have to explore if it leads to a group which is fully reducible and hence finite.

Commutation Rules for Real matrices :
Let A =
a    b
c    d
and B=
e    f
g   h
Rule 1 :  b/f = c/g = k (say)
Rule 2 : (d-a)/b = (h-e)/f   or k= b/f = c/g = (d-a)/(h-e)
In order to modify the matrix B so that it commutes with matrix A, we can keep any 2 elements of B unchanged and then modify the other 2 elements so that Rule 1 and 2 are complied with.
Case 1: e, f are kept intact.
               g = c/k
               h = d/k + (e - a/k)
Case 2: h, f kept intact
            g= c/k
            e= a/k + (h- d/k)
Case 3: g, h kept intact
            f = b/k
            e =a/k +(h-d/k)
Case 4: e, g kept intact
            f =b/k
            h =d/k +(e- a/k)
Interchange occurs between
b <->c
a <->d
e <->h
g <->f
Value of k in all cases above is pre fixed by choice of 2 intact elements.
Case 5: e, h fixed
          f = b/k
          g = c/k
Case 6: g, f kept intact
             not possible to find e, h.
We know that if A, B are 2 commutative  matrices i.e. [A, B ] =0 , they both can be diagonalized simultaneously by any one of the similarity matrices Sa or Sb. In Quantum mechanics, if 2 operators are commutative, the corresponding observables which are their corresponding eigen values can be measured simultaneously with 100% accuracy as the operators do not influence each other and hence Heisenberg's uncertainty is not applicable here.           

Let X =(anti-trA)/2  and

     Y=√{[(anti-trA)/2]² + bc]

then Sa =                    Sa1                                                                                       Sa2
         (Sa1/b)*[(anti-trA)/2 + √[(anti-trA)/2]² + bc ]                  (Sa2/b)*[(anti-trA)/2 - √[(anti-trA)/2]² + bc ]
 

 =                                Sa1                                                                                       Sa2
                        (Sa1/b)*[X + Y ]                                                                    (Sa2/b)*[X - Y ]

For Sa to be orthogonal matrix,

Sa1= cosθ  .....(1)

Sa2=-sinθ ......(2)

(Sa2/b)*[X - Y ] =cosθ .or or -cotθ =(X-Y)/b.......(3)

(Sa1/b)*[X + Y ] =sinθ  or tanθ =(X+Y)/b.......    (4)

(3) * (4)  Y² -X² = b² or bc=b²   or b=c

Matrix A has to be a symmetric matrix.

 tanθ =(X+Y)/b  or cosθ = b   /   √ [ b² +(X+Y)² ]

                               sinθ = (X+Y)  /  √ [ b² +(X+Y)² ]

Sa =   cosθ               - sinθ

         sinθ               cosθ

If Sa is to be

Sa =   cosθ                sinθ

         sinθ             cosθ

then X² -Y² = b² or - bc=b²   or b=- c

 tanθ =(X+Y)/b = anti-tr / 2b  + √[(anti-trA)/2]² + bc ] /b = anti-tr / 2b  + √[(anti-trA)/2]² - b² ] /b

or  tanθ  = anti-trA / 2b +  √[(anti-trA)/2b]² - 1 ]

if A=

a   b

c  d

and S=

e    f

g   h

where S is the similarity matrix

S^-1AS=

(h/Δ) *(ae+bg) - (f/Δ)*(ce+dg)               (h/Δ) *(af+bh) - (f/Δ)*(cf+dh) = A' =  a11       a12   =λ₁        0

(-g/Δ) *(ae+bg) + (feΔ)*(ce+dg)           (-g/Δ) *(af+bh) - (f/Δ)*(cf+dh)            a21        a22       0       λ₂

_Equating   (-g/Δ) *(ae+bg) + (feΔ)*(ce+dg)  =  (h/Δ) *(af+bh) - (f/Δ)*(cf+dh)  = 0

or,   g = (e/b)[(d-a)/2  ± √{(d-a)² /4  + bc}] =(e/b)(λ₁-a )

       h = (f/b)[(d-a)/2  ± √{(d-a)² /4  + bc}]=  (f/b)(λ₂-a)

Equating the diagonal elements, we get an identity and hence this equation does not yield any useful result.

we have to arbitrarily choose e,f.