Linear Algebra (matrix) Proof. Super stuck:?

Question

I need to prove that if A and B are matrices such AB and BA exist, and AB is symmetric (so AB = (AB)transpose ), show that AB=Ba.

So far I can prove that AB and BA have the same size, but I can't figure out how to prove that they have the same entries.
Proof of size:
A is m*n, B is n*p. So AB is m*p, but since BA exists too m must = p. So now A is m*n, and B is n*m. This leads to AB being m*m and BA being n*n. And since AB=BA m must = n so all of the mentioned matrices must be n*n in size.
now i just need to prove that the entries of AB and BA are the same.

It's not a homework question. Its a problem in my book, but it doesn't have all the answers in the back. I'm just trying to do some extra questions for practice. I think the answer should just be explained, not done by example. Like how I got the fact that they are the same size.

Crap. I didn't see the bit above the question that said "assume that A and B are symmetric". The question is a whole lot easier now.
Thanks for the help though!

Answer 1

Assume A is an mXn and B is an nXp since AB exists, then AB is an mXp and BA would be a nXpXmXn. But the only way BA can be that is if p = m, in which case nXpXmXn = nXmXmXn = nXn; so that the BA is symmetric in n. Also, since m=p, AB is an mXm or pXp...also symmetric, but in m or p.

I don't see AB = BA here. In fact, I see generally that AB <> BA. Further, I don't see where the transpose of either term has any bearing on this since both matrix multiplications are symmetric, but in different dimensions.

Sorry I can't help; perhaps I misunderstood the problem

Answer 2

The original question is false. Let A be the 2x2 matrix (1,0; 0,0) [I'm listing the first row, then the second row.], and B the 2x2 matrix (0,0;1,0). Then AB=(0,0;0,0) which is symmetric, but not equal to BA=(0,0;1,0). "Random" matrices rarely commute. However if AB is nonzero diagonal then AB=BA!

Also something is wrong with your proof that they have the same size, since A could be 2x1 and B could be 1x2. AB is then a 2x2 matrix, and BA is a 1x1 matrix. Your flaw is when you said AB=BA which is what you were trying to prove!

Answer 3

Transpose is written as ' (promounced prime)

the rule is that (AB)' = B'A'

so your result only works if A and B are symmetric,

I am not sure if that follows from AB being symmetric

Answer 4

Well this seems like a homework assignment so I do not want to give the answer away but you could try writing out the matricies like this
[ i1j1 i1j2 ... i1jn]
[ i2j1 i2j2 ... i2jn]
... ... ... ... ...
[imj1 imj2 ... imjn]

Then multiplying them out

Also what kind of matrices are symetric? Can a rectangle matrix be simetric?