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.
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.
Diagonalization Solution
