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Matrix Product Poser (Posted on 2016-03-19) Difficulty: 3 of 5
A is a 3x2 matrix and B is a 2x3 matrix, and:
AB = 
⌈ 8  2 -2⌉
| 2  5  4|
⌊-2  4  5⌋
Find the product BA.

No Solution Yet Submitted by K Sengupta    
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Solution Diagonalization Solution | Comment 1 of 3
Let M=A*B.  M is symmetric and all symmetric matricies are diagonalizable.  The rank of M is 2 and its nullity is 1.  In general any diagonalizable mxm matrix M with rank r can be expressed as a product of a pair of matricies A (mxr) and B (rxm). 

Let P be a matrix that diagonalizes M and let D be the corresponding diagonal matrix.  The elements on the diagonal of D are the eigenvalues of M.  There will be r nonzero eigenvalues.  Arrange these so that the 0's are in the lower right corner.  

Then D can be factored into A' and B'. A' is an rxr identity matrix, augmented below with a (m-r)xr zero matrix; B' is a diagonal matrix whose diagonal elements are the nonzero values of D, augmented with a rx(m-r) zero matrix to the right.

Then D = A'*B' implies M = P*A'*B'*P^-1.  Let A = P*A' and let B = B'*P^-1.  Then the product B*A = (B'*P^-1)*(P*A') = B'*A', which is a rxr diagonal matrix with the nonzero eigenvalues of M on the diagonal.

For the specified M: its nonzero eigenvalues are 9 and 9.

Then B*A = [ 9  0 ]
           [ 0  9 ]
To check this I went through the full process with:

    [ 2  0  1 ]      [ 9  0  0 ]        [ 4/9  1/9 -1/9 ]
P = [ 0  1 -2 ]  D = [ 0  9  0 ] P^-1 = [ 2/9  5/9  4/9 ]
    [-1  1  2 ]      [ 0  0  0 ]        [ 1/9 -2/9  2/9 ]

    [ 2  0 ]      [ 4  1 -1 ]
A = [ 0  1 ]  B = [ 2  5  4 ]
    [-1  1 ]

Everything checks out with this specific valuation of A and B.


  Posted by Brian Smith on 2016-03-19 10:57:04
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