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.

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**