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 -Δ (tr/2)2 -Δ 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 λ1/λ2 +i λ1/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 B MatrixSB (e) (f) (a6) (b6) (g) (h) (c6) (d6) Δ(det): tr: Δ(det): tr: (tr/2)2 -Δ (tr/2)2 -Δ anti-tr: anti-tr/2b anti-tr: anti-tr/2b √[(tr/2)2 -Δ] +i √[(tr/2)2 -Δ] +i λ1 +i λ1 +i λ2 +i λ2 +i λ1/λ2 +i λ1/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 -Δ (tr/2)2 -Δ 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 (1) & (3) b=c ; d/a = u/y (a1) (b1) (u1r) + i (u1i) (v1) (u1ar) + i (u1ai) (c1) (d1) (w1) (y1r) +i (y1ar) +i(y1ai) Find Similarity matrix satisfying condition (1) & (3) b=c ; v=w (a2) (b2) (u2) (v2r) +   i   (v2i) (v2ar) + i (v2ai) (c2) (d2) (w2r) +   i   (w2i) (y2) (w2ar) + i (w2ai) Find Similarity matrix satisfying STAS=Ka AS=(ST)-1Ka ,   b=c (a5) (b5) (u5) (v5) (c5) (d5) (w5) (y5) Matrix L=SAS-1 Matrix La=S-1AS (L11) (L12) (La11) (La12) (L21) (L22) (La21) (La22) Δ(det): tr: Δ(det): tr: (tr/2)2 -Δ (tr/2)2 -Δ anti-tr: anti-tr/2L12: anti-tr: anti-tr/2La12: √[(tr/2)2 -Δ] +i √[(tr/2)2 -Δ] +i Find Similarity matrix SA satisfying condition S-1AS=La or AS=SLa ....(4) (a3) (b3) (u3) (v3) (c3) (d3) (w3) (y3) Find Similarity matrix satisfying condition SAS-1=L or SA=LS ....(5) (a4) (b4) (u4) (v4) (c4) (d4) (w4) (y4)

 * In case of an orthogonal matrix S,  ST=S-1 . Matrix A can be diagonalized by applying transformation S-1AS =STAS .* 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 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 STAS =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=STAS = 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=SAST = 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=(ST)-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, STAS = Ka which is diagonal matrix with diagonal values being eigen values. But in normal cases       S-1AS = Ka  which implies that ST =S-1  which is the case for orthogonal matrices. (a)Similarity matrix for a square matrix A satisfies the equation S-1AS=La where La is a diagonal matrix having elements λ1 and λ2 which are the eigen values of A. We can also write AS=LaS..(6) or  a  b   * u    v  = u    v   *  λ1     0   or  au+bw    av+by   =uλ1         vλ2                                                                              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) Commutation product of 2 Real matrices : Let A = a    b c    d and B= e    f g   h[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=0Case 2: When b=c=0   [A,B] = 0             -f(d-a)  and determinant = fg(d-a)2                                                 g(d-a)           0Case 3: When f=g=0  [A,B] = 0             b(h-e)  and determinant = bc(h-e)2                                               -c(h-e)           0Case 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 Suppose D =  x     X    is a block matrix where X,Y are also matrices and x,y are scalar numbers.                       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. Similarity Matrix: For 2x2 matrices, to find out similarity matrix, We can arbitrarily take any 2 matrix elements and then compute the other 2 elements . Sa =  Sa1   Sa2 Similarity matrix for A          Sa3   Sa4 Sb =  Sb1   Sb2 Similarity matrix for B          Sb3   Sb4 In this example, We take (Sa1, Sa2) & (Sb1, Sb2) as arbitrary. Then, it can be proved that Sa =                    Sa1                                                                                       Sa2          (Sa1/b)*[(anti-trA)/2 + √{[(anti-trA)/2]2 + bc ]                  (Sa2/b)*[(anti-trA)/2 - √{[(anti-trA)/2]2 + bc ] Sb =                    Sb1                                                                                       Sb2          (Sb1/f)*[(anti-trB)/2 + √{[(anti-trB)/2]2 + fg ]                  (Sb2/f)*[(anti-trB)/2 - √{[(anti-trB)/2]2 + fg ] If [A,B] = 0 then Sb =                    Sb1                                                                                       Sb2          (Sb1/b)*[(anti-trA)/2 + √{[(anti-trA)/2]2 + bc ]                  (Sb2/b)*[(anti-trA)/2 - √{[(anti-trA)/2]2 + bc ] When Similarity Matrix becomes an Orthogonal Matrix ?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=cMatrix 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               cosθIf 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 ] 