Question

5. A matrix is symmetric if it is equal to its transpose. Let \[A\] be an mxn matrix.
(a) Explain why the matrix multiplications \[A^{T}A\] and \[AA^{T}\] are always possible.
(b) Show that \[A^{T}A\] and \[AA^{T}\] are both symmetric

Answer 1

For (a) I have

We have \[A_{mxn}\] by definition, then \[A^{T}_{nxm}\]

In order to multply two matrices, the number of columns of the left matrix must equal the number of rows of the right matrix. For some m x n matrix A, the number of columns of A is the number of rows of A^{T}, so multiplication is always possible. For A^{T}A, where A^T nxm, A mxn, the same argument applies.

Answer 2

For (b), I have \[A^{T^{T}} = A ; (A^{T}A)^{T} = AA^{T},\] and for \[(AA^{T})^{T} = A^{T}A^{T^{T}} = A^{T}A. \] This doesn't seem like a full or formal proof, and I think to bring it all together I'd have to show that \[(A^{T}A)^{T} = A^{T}A\], but I'm not certain how to do that because matrix multiplication isn't commutative.

Answer 3

one moment, checking my linear algebra book

Answer 4

Ok!

Answer 5

Got it. Answer:
Assuming A is symmetric,
\[A=A^{T};A^{T}A=A^{2};AA^{T}=A^{2};A^{T}A=AA^{T}\]

Answer 6

why are you assuming A is symmetric?

Answer 7

If A isn't symmetric can't commute (A^T)A to AA^T. Only way I can get it to work, I just made the assumption that it was left out of the question because professor's are fallible. Is there another way to do it?

Answer 8

for any A and B where AB is defined we have

\((AB)^T=B^TA^T\)

Answer 9

So need \[AA^{T}=(AA^{T})^{T};(AA^{T})^{T}=A^{T}A \neq AA^{T},\] implicitly anyway. Need commutativity still, right?

Answer 10

\[(AA^T)^T=A^{T^T}A^T=AA^T\]

therefore \(AA^T\) is symmetric

Answer 11

Thank you!!! I don't know why I didn't see that when you gave the relation the first time.