Is it necessary $A^TA =I$ if $AA^T=I$ where A is a square matrix?

Question

Suppose if $AA^T=I$ where $A$ is a square matrix, is it necessary that $A^TA=I$?

I tried to google it but didn't find anything confirmatory.

In the same way if $AA^{-1}=I$, is it necessary that $A^{-1}A=I$?

My attempt it as follows:-

$$A=\begin{bmatrix} a&b&c\\ d&e&f\\ g&h&i \end{bmatrix}$$

$$AA^T=\begin{bmatrix} a&b&c\\ d&e&f\\ g&h&i \end{bmatrix}\begin{bmatrix} a&d&g\\ b&e&h\\ c&f&i \end{bmatrix}$$

$$AA^T=\begin{bmatrix} a^2+b^2+c^2&ad+be+cf&ag+bh+ci\\ da+eb+fc&d^2+e^2+f^2&dg+eh+fi\\ ga+hb+ic&gd+he+if&g^2+h^2+i^2 \end{bmatrix}$$

$$a^2+b^2+c^2=1$$ $$d^2+e^2+f^2=1$$ $$g^2+h^2+i^2=1$$ $$ad+be+cf=0$$ $$ag+bh+ci=0$$ $$da+eb+fc=0$$ $$dg+eh+fi=0$$ $$ga+hb+ic=0$$ $$gd+he+if=0$$

Now if we calculate $A^TA$, it would be

$$A^TA=\begin{bmatrix} a^2+d^2+g^2&ab+de+gh&agc+df+gi\\ ba+ed+hg&b^2+e^2+h^2&bc+ef+hi\\ ca+fd+ig&cb+ef+ih&c^2+f^2+i^2 \end{bmatrix}$$

Now can one say $a^2+d^2+g^2=1$ from the previous equations? I think not necessary. So by this exercise I doubt $A^TA=I$

I am not getting any idea for $AA^{-1}=I$

yes I saw that, but nobody said clearly that it is true or not? I just want the confirmation. — user3290550, Dec 03 '19 at 00:05
There are more than 20 proofs of this facts in the link already. What sort of confirmation are you asking for besides the proof? — Arctic Char, Dec 03 '19 at 00:11

score 2 · Answer 1 · answered Dec 03 '19 at 00:08

2

Yes, because inverses are unique.

answered Dec 03 '19 at 00:08

SescoMath

score 0 · Answer 2 · answered Dec 03 '19 at 00:16

0

Yes. To see this note that since $I=A^TA$ we have $A=A(A^TA)=(AA^T)A$ so $AA^T=I$.

answered Dec 03 '19 at 00:16

CyclotomicField

score -3 · Answer 3 · answered Dec 03 '19 at 00:23

-3

In addition to all this $I=I^T=(AA^T)^T =A^T(A^T)^T=A^TA $.

answered Dec 03 '19 at 00:23

Anonmath101

3 Answers3