Eigenvalue problem 

                      | A11  A12  A13  |
Assume a matrix A  =  | A21  A22  A23  |
                      | A31  A32  A33  |
   | 4  3  7 |
A= | 8  9  5 |   Example
   | 6  2  4 |

The goal is to find the groups 
A*CC1 = CC1*B1     A*CC2 = CC2*B2     A*CC2 = CC2*B2

      | C11  |          | C21  |          |  C31   |
CC1 = | C12  |    CC2 = | C22  |    CC3 = |  C32   |
      | C13  |          | C23  |          |  C33   |

                       | A11-X  A12    A13    |
Assume a matrix AX  =  | A21    A22-X  A23    | = 0
                       | A31    A32    A33-X  |

Compute determinate of AX = 0
Solve for X. The solutions will be C1, C2 and C3.
Example  -(X^3-17*X^2+12*X+168)=0
The three solutions of X are 
B1=  15.5308476265
B2=  -2.63540657687
B3=   4.10455895037

We will solve for one case
A*C1 = C1*B1   Multiply both sides by (INVERSE A)
A*(INVERSE A)= IDENTITY MATRIX  
This gives C1=(INVERSE A)*C1*B1
The values C11, C12, C13 are to be determined.
Assume one member of each vector=1, as any arbitrary value may be assumed.
(INVERSE A)*C1*B1-C1=0
Let (INVERSE A)*B1=U1

    | U11 U12 U13 |
U1= | U21 U22 U23 |
    | U31 U32 U33 |

| C11  |   | U11 U12 U13 |    | C11  |
| C12  | * | U21 U22 U23 | -  | C12  | = 0
| C13  |   | U31 U32 U33 |    | C13  |   
  
Since C12=1, this leaves two unknowns, C11 and C13.

C13 * U13 + C12 * U12 + C11 * U11 = C11
C13 * U33 + C12 * U32 + C11 * U31 = C13

Since C12=1, this leaves two unknowns, C11 and C13.
solve. for C11 and C13
 
     | 0.534216183955  |      | -2.91144642594 |       | -.34089056339162  |
CC1= |       1.0       | CC2= |     1.0        |  CC3= | 1.0               |
     | 0.451423630971  |      | 2.33123296612  |       |  -4.3366330849979 |

               |  8.29683015262 |                 | 7.67284505911  |   
A*CC1=CC1*B1 = | 15.5308476265  | A*CC2=CC2*B2 =  | -2.63540657687 |
               |  7.01099162761 |                 | -6.14374669113 |
                        
                                  |-1.39920541307 |   
                   A*CC3=CC3*B3 = | 4,10455895037 |
                                  |-1.77999661435 |

Some simultaeous equations give bad solutions. Check for validity.
========
TI-89 method
approximate mode

   | 4  3  7 |
a= | 8  9  5 |   Example
   | 6  2  4 |

eigVc(a)

|  -.437768863071     -.753968114055   .298490962544  |
|  -.819460129099      .258966851438  -.875621077844  |  =c
|  -.369923666914      .603712061205   .37972473361   |


eigVl(a)
{ 15.5308476265  -2.63540657687   4.10455895037 }  =b

subMat(c,1,1,3,1) -> cc1

|  -.437768863071  |
|  -.819460129099  |  = cc1
|  -.369923666914  |
----------------------------

subMat(c,1,2,3,2) -> cc2

| -.753968114055   |
|  .258966851438   |  = cc2
|  .603712061205   |
--------------------------
subMat(c,1,3,3,3) -> cc3

|  .298490962544  |
| -.875621077844  |  = cc3
|  .37972473361   |
---------------------------
product(mid(c,1,1)) ->b1 
15.5308476265 = b1

--------------------------
product(mid(c,2,1)) ->b2 
-2.63540657687 = b2
--------------------------
product(mid(c,3,1)) ->b3 
4.10455895037 = b3
-------------------------
a*cc1= b1*cc1  =

|  -6.79892150798  |
| -12.726910401    |
|  -5.74522810428  |

--------------------------
a*cc2=b2*cc2   =

| 1.98701252653  |
| -.692482943471 |
| -1.59102673663 |
--------------------------
a*cc3=b3*cc3   =

| 1.22517375191  |
| -3.5940383322  |
| 1.55860255401  |
========================