Let A and B be symmetric nxn matrices. Prove that AB = BA if and only if AB is also symmetric?

Question

I get so close. I feel like the proof involves transposes and inverses but I can't put it all together.

Answer 1

Let us denote transpose of a matrix M as M'

A is symetric => A = A'
B is symetric => B = B'

Now AB = BA
take transpose of both and use (AB)' = B' A'

(AB)' = (BA)'
= A'B'
= AB
so (AB)' = AB so AB is symetric

Now to prove the other way
(AB)' = AB = A'B' = (BA)' = (B'A')' = ((AB)')' = AB
so we have proved other way also
so iff holds

Answer 2

If you have been studying matrices and their transposes, then you probably already have seen the theorem that says

transpose of AB = (transpose of B)(transpose of A) for any n x n matrices A and B.

Now if we know that A and B are symmetric, then by definition transpose of A = A and transpose of B = B, and so by the theorem above, transpose of AB = BA.

Now we are ready to answer your question.

First let's assume that AB = BA. Then, using our previous result. transpose of AB = BA = AB, and so by definition AB is symmetric.

Now we assume that AB is symmetric, i.e. that AB = transpose of AB. Then using the previous result again, we see that AB = transpose of AB = BA.

That should do it. Good luck!

Answer 3

let C=AB and D = BA

then Cij = AinBnj = AniBjn = BjnAni = Dji

this shows that if A ans B are symmetric then AB is symmetric too

Answer 4

A is symmetric means a(i,j) = a(j,i)
B is symmetric means b(i,j) = b(j,i)

Write c(i,j) for the elements of AB and d(i,j) for those of BA,

c(i,j) = SUM a(i,k) b(k,j)
d(i,j) = SUM b(i,k) a(k,j)

Note now that
c(i,j) = SUM a(i,k) b(k,j) = SUM b(j,k) a(k,i)
... = SUM d(j,i)

so AB and BA are transposes of each other.

If AB = BA, then AB equals its own transpose, therefore it is symmetric.

Answer 5

I tried SO HARD to understand what that siad, I break it down and read it slowly... no luck. My math class is just learning that stuff. Good luck! Peace out! Bye. Luv/Kodi