When is it it true that $AA^T=A^TA$?

Question

I know for orthogonal matrices, $A^TA=I$. Does this also mean that $AA^T = I$?

I think yes, because $(A^TA)^{-1} = I^{-1} = I$

Is this correct?

Answer 1

Whereas it is certainly true that $A^TA = I$ implies $(A^TA)^{-1} = I^{-1} = I$, the argument suggested by our OP Joshua Bone then needs to be completed with the assertion that

$A^{-1}(A^T)^{-1} = (A^TA)^{-1} = I \Longrightarrow A = (A^T)^{-1} \Longrightarrow AA^T = I, \tag 0$

which requires the invertibility of $A$ and $A^T$; thus, I reason as follows:

The hypothesis

$A^TA = I \tag 1$

implies that $A$ is nonsingular; for if there is a vector

$w \ne 0 \tag 2$

with

$Aw = 0, \tag 3$

then we may write

$w = Iw = A^TAw = A^T(0) = 0, \tag 4$

contradicting (2), the nonsingularity of $A$ follows. This in turn implies the existence of $A^{-1}$, whence from (1)

$A^T = A^TI = A^T(AA^{-1}) = (A^TA)A^{-1} = IA^{-1} = A^{-1}; \tag 5$

we now left multiply by $A$ and obtain

$AA^T = AA^{-1} = I, \tag 6$

as desired.

Answer 2

Sure it does, as long as $A$ is supposed to be square since $$AA^T=I\iff A^T=A^{-1}$$

Answer 3

A matrix commuting with its own conjugate transpose (which, in the real case, is just the transpose) is called normal. The spectral theorem states a square matrix is normal iff unitarily similar to a diagonal matrix.