Can we conclude from $MM^T=I$ that $M^TM=I $ over $\mathbb F_2$

Question

Let $M$ be nonsingular real matrix, dimension $n$. If we know that $$MM^T=I,$$ where $I$ is identical matrix. Is it true that $M^TM=I$ too. And it is all over $\mathbb F_2$ field, which is field that consists only of $0,1$.

My first thoughts was that it is not necessary true, but then:

• If we multiply our main equation by $M$ on the right we get: $$MM^TM=M $$
• If we multiply our main equation by $M^T$ on the left, we have: $$M^TMM^T=M^T $$

So, can we conclude from here, that $M^T=I?$ Or there can exist other matrix $A\neq I$ such that $AM^T=M^T$ and $MA=M?$

Answer 1

Hint: What does $MM^T=I$ tell you about $M^{-1}$?

Answer 2

Yes. The right inverse of a matrix is equal to the left inverse. If $MM^T=I$, $M^T$ is the right inverse of $M$. So it is also the left inverse and $M^TM=I$.

Answer 3

Yes, because for square matrices, a right inverse is automatically a left inverse.