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 Matrix Product Poser (Posted on 2016-03-19)
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 No Rating

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 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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