AA' A = A
(AA')2 = A
AA'A = $$I$$
AA'A = A'
Answer is Right!
Answer is Wrong!
The correct answer is $\boxed{\text{C}}$.
Let $A$ be an $m\times n$ full rank matrix with $m>n$. Then, $A^T A$ is an invertible matrix. Therefore, $(A^T A)^{-1}$ exists.
We know that $A^T A A = A^T$. Therefore,
\begin{align}
AA’ A &= A (A^T A)^{-1} A^T \
&= A (A^T A)^{-1} A \
&= I.
\end{align}
Option A is not true because $AA’ A$ is an identity matrix, not $A$.
Option B is not true because $(AA’)^2 = AA’ A A’ = AA’$.
Option D is not true because $AA’ A$ is an identity matrix, not $A’$.