Show that A’A and

We must understand,

In linear algebra, a symmetric matrix is a square matrix that is equal to its transpose. Formally, because equal matrices have equal dimensions, only square matrices can be symmetric.

And we know that, transpose of AB is given by

(AB)’ = B’A’

Using this result, take transpose of A’A.

Transpose of A’A = (A’A)^T = (A’A)’

Using, transpose of A’A = (A’A)’

⇒ (A’A)’ = A’(A’)’

And also,

(A’)’ = A

So,

(A’A)’ = A’A

Since, (A’A)’ = A’A

This means, A’A is symmetric matrix for any matrix A.

Now, take transpose of AA’.

Transpose of AA’ = (AA’)’

⇒ (AA’)’ = (A’)’A’ [∵ (AB)’ = B’A’]

⇒ (AA’)’ = AA’ [∵ (A’)’ = A]

Since, (AA’)’ = AA’

This means, AA’ is symmetric matrix for any matrix A.

Thus, A’A and AA’ are symmetric matrix for any matrix A.