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Rectangular Matrix Product Poser (Posted on 2016-04-16) Difficulty: 4 of 5
Let A and B be the two matricies:
      [ 4  2 -10 ]       [ 0  3 ]
  A = [-1  1   5 ]   B = [-2  5 ]
      [-1 -2  -2 ]  
Find a 3x2 matrix C with rank 2 such that A*C = C*B.

  Submitted by Brian Smith    
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Solution: (Hide)
Matricies A and B are not the same size. If they were the same size (both 3x3) this would imply that A and B were similar matricies. In particular they would have similar diagonalization/Jordan form and have identical eigenvalues.

The eigenvalues of A are -2, 2, and 3. The eigenvalues of B are 2 and 3. The eigenvalues of B are a subset of A. Pad out B to create B' including the third eigenvalue -2:
      [ 0  3  0 ]
 B' = [-2  5  0 ]
      [ 0  0 -2 ]
Now A and B' are similar matricies. They can both be diagonalized with the same diagonal matrix D. Let A = P*D*P^-1 and B' = Q*D*Q^-1. Then one diagonalization is:
     [ 2  0  0 ]      [ 14  2  2 ]      [ 3  1  0 ]
 D = [ 0  3  0 ]  P = [ -9 -1 -1 ]  Q = [ 2  1  0 ]
     [ 0  0 -2 ]      [  1  0  1 ]      [ 0  0  1 ]
Rearrange the equations for diagonalization to yield D = P^-1*A*P = Q^-1*B'*Q. Left multiply each side by P and right multiply each side by Q^-1 to get A*P*Q^-1 = P*Q^-1*B'. Then C'=P*Q^-1 is a matrix which solves A*C' = C'*B'.
      [ 10 -8  2 ]
 C' = [ -7  6 -1 ]
      [  1 -1  1 ]
Finally C' can be reduced to create C be removing the rightmost column (C will have rank 2 because C' has rank 3 from being invertible):
     [ 10 -8 ]
 C = [ -7  6 ]
     [  1 -1 ]
A*C = C*B can be verified by direct multiplication:
            [ 16 -10 ]
A*C = C*B = [-12   9 ]
            [  2  -2 ]

Comments: ( You must be logged in to post comments.)
  Subject Author Date
re: ComplexLove Guru2016-04-20 06:35:06
Complexarmando2016-04-19 15:50:22
possible solution enlargedarmando2016-04-18 09:43:30
possible solutionarmando2016-04-17 12:08:10
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