All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars
 perplexus dot info

 Rectangular Matrix Product Poser (Posted on 2016-04-16)
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 No Rating 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 ] ```

 Subject Author Date re: Complex Love Guru 2016-04-20 06:35:06 Complex armando 2016-04-19 15:50:22 possible solution enlarged armando 2016-04-18 09:43:30 possible solution armando 2016-04-17 12:08:10

 Search: Search body:
Forums (0)