Similarity Transformation

of

3x3 Real Matrix -B and A

A-1BA=B'   and B-1AB=A'

&

Diagonalization of B, A

 Matrix B det: Trace: Matrix A det: Trace: adj.B=D det: Trace: adj.A=C det: Trace: B-1=F det: Trace: A-1=E det: Trace: BA=G det: Trace: AB=H det: Trace: A-1BA=J det: Trace: B-1AB=K det: Trace: Characterstic Equn |B-λ1I| =0 λ3 + (A1) λ2  + (B1) λ + C1 = 0 λ3 + λ2  + λ + = 0 λ1: λ2: λ3: *manually put values of λ1,λ2,λ3 below  and press submit at topmost left. B' 0 0 0 0 0 0 * Now after finding the diagonal matrix, matrix A has to be found out.Equations are(b11-λ1) a1+ b12a4        +b13a7=0             ---(1)                        b21a1        +(b22-λ1) a4+b23a7=0          .... (2)                        b31a1        +b32a4         +(b33-λ1) a7=0      ......(3) Solving, a4=a1x1   where  x1=[b21b13-(b11-λ1)b23] / [b21b23-(b22-λ1)b13]            a7=a1x2  where    x2=[b12b31-(b11-λ1)b32] / [b13b32-(b33-λ1)b12] Out of 3 unknowns, value of 1 has to be specified and then the value of other 2 can be determined. Equations are(b11-λ2) a2+ b12a5        +b13a8=0             ---(4)                        b21a2        +(b22-λ2) a5+b23a8=0          .... (5)                        b31a2        +b32a5         +(b33-λ2) a8=0......(6) Solving, a5=a2y1 where  y1=[b13b21-b23(b11-λ2)] / [b12b23-b13(b22-λ2)]             a8=a2y2 where y2= [b31b12-b32( b11-λ2)]/ [b13b32-b12(b33-λ2)] Out of 3 unknowns, value of 1 has to be specified and then the value of other 2 can be determined.Equations are(b11-λ3) a3+ b12a6        +b13a9=0             ---(7)                        b21a3      +(b22-λ3) a6+b23a9=0          .... (8)                        b31a3        +b32a6      +(b33-λ3) a9=0......(9) Solving, a6=a3z1 where z1=[b13b21-b23(b11-λ3)] / [b23b12-b13(b22-λ3)]              a9=a3z2 where z2=[b12b31-b32(b11-λ3)] / [b13b32-b12(b33-λ3)] Out of 3 unknowns, value of 1 has to be specified and then the value of other 2 can be determined.Thus (a1,a4,a7) -> one has to be arbitrarily fixed           (a2,a5,a8) -> one has to be arbitrarily fixed           (a3,a6,a9) -> one has to be arbitrarily fixed By convention, we can fix a1,a2,a3 and then compute the others .The arbitrarily fixed ones are called the basis if they are treated as unity. calculate a1 to a9 a1 a4 a7 a2 a5 a8 a3 a6 a9 λ1: λ2: λ3: characteristic equation |A-λ1I| =0 λ3 + (A2) λ2  + (B2) λ + C2 = 0 λ3 + λ2  + λ + = 0 λ1: λ2: λ3: *manually put values of λ1,λ2,λ3 below  and press submit at topmost left. A' 0 0 0 0 0 0 calculate b1 to b9 b1 b4 b7 b2 b5 b8 b3 b6 b9 λ1: λ2: λ3:

 * Diagonalization of the matrix: applicable for non-singular matrices. A-1BA=B' or BA=AB' where B' is a diagonal matrix whose diagonal elements are λ1 ,λ2, λ3. B= b1 b2 b3   and  A = a1 a2 a3 then [ b1 -λ1   b2       b3        * [a1                [ b1 -λ2   b2       b3        * [a2                  [ b1 -λ3   b2       b3        * [a3        b4 b5 b6                  a4 a5 a6            b4      b5-λ1     b6            a4   =0   ;       b4      b5-λ2   b6            a5   =0                b4     b5-λ3     b6            a6    = 0         b7 b8 b9                  a7 a8 a9            b7       b8        b9-λ1]       a7 ]              b7       b8        b9-λ2]       a8 ]                   b7       b8        b9-λ3]       a9 ]* The general characteristic equation is      [ b1 -λ   b2       b3        * [x                                                                       b4      b5-λ     b6            y   =0   where there are 3 solutions to λ and for each λ, (x,y,z) are (a1,a4,a7), (a2,a5,a8) & (a3,a6,a9) respectively .                                                                        b7       b8      b9-λ]         z ] * since these are homogeneous equations , the matrix is singular and hence out of x,y,z one has to specify value of either x,y,z and then others can be found out.