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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 Orthogonal Diagonalization | Comment 2 of 3 |
In my earlier post I noted that the matrix was symmetric and therefore diagonalizable.  But there is a much stronger claim; it is orthogonally diagonalizable.  

This combined with the fact that all the nonzero eigenvalues are positive reals implies there exists matricies A and B such that they are transpose of each other: B = A^T.

Let M=A*B.  The orthogonal diagonalization of M is M = P*D*P^T with P^T = P^-1.  D would then be factored into an mxr A' with the square roots of the eigenvalues on the diagonal and B' would be its transpose.

Then M = P*A'*A'^T*P^T.  Let A = P*A' and B=A'^T*P^T.  B is the transpose of A from the property (X*Y)^T = Y^T*X^T.

B*A will give the same rxr matrix as before with the nonzero eigenvalues of M on its diagonal.

For the given matrix M:
    [ 2/3 -2/3  1/3 ]      [ 9  0  0 ]      [ 2 -2 ]        [ 9  0 ]
P = [ 2/3  1/3 -2/3 ]  D = [ 0  9  0 ] A = [ 2  1 ] B*A = [ 0 9 ]
    [ 1/3  2/3  2/3 ]      [ 0  0  0 ]     [ 1  2 ]
These matricies check out with B=A^T.

Edited on March 19, 2016, 3:00 pm
  Posted by Brian Smith on 2016-03-19 14:58:46

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