# Does $A^TA=I$ imply $AA^T=I$?

Let $$m,n\in\mathbb R^{m\times n}$$ with $$A^TA=I_n\tag1.$$ I wonder whether this implies that $$AA^T=I_m\tag2$$ or if we can show it at least in the case $$m=n$$.

EDIT: I was hoping for a proof which directly shows the orthonormality of the columns implies the orthonormality of the rows.

If $$(e_1,\ldots,e_n)$$ and $$(f_1,\ldots,f_m)$$ denote the standard bases of $$\mathbb R^n$$ and $$\mathbb R^m$$, then $$(1)$$ is equivalent to saying that the columns $$Ae_1,\ldots,Ae_n$$ are orthonormal and $$(2)$$ is equivalent to saying that the rows $$A^T f_1,\ldots,A^T f_m$$ are orthonormal.

An indirect argument is somehow clear to me. Multiplying $$(1)$$ by $$A$$ and $$A^T$$ from the left and right, respectively, yields $$(A^TA)^2=A^TA\tag3,$$ which means that $$A^TA$$ is a projection. In the same way, multiplying $$(1)$$ only by $$A$$ from the left yields $$(I_m-AA^T)A=0\tag4.$$ Now, by $$(1)$$, $$A^T\mathcal R(A)=\mathcal R(A^T A)=\mathbb R^n\tag4$$ and hence $$n=\dim A^T\mathcal R(A)\le\dim\mathcal R(A)\le n\tag5$$ by the rank-nullity theorem. This implies that $$n\le m$$ by the way.

(a) How can we conclude that, if $$m=n$$, then $$I_m-AA^T=0$$?

(b) Can we show the orthonormality of $$A^T f_1,\ldots,A^T f_m$$ directly?

• Does this answer your question? Does $AA^T=I$ imply that $A^TA=I$? - and conversely. May 29, 2020 at 10:59
• @DietrichBurde No, since the answer basically restates the question. ($A^T$ is the left-inverse implies $A^T$ is the inverse, is the same question as asking for $A^T$ is the left-orthogonal implies $A^T$ is orthogonal). May 29, 2020 at 11:02
• Both follows trivially from the fact the left-inverse in the group $O(n)$ is right-inverse - which is the duplicate. Well, for $m=n$, I am sorry. Otherwise it is false. May 29, 2020 at 11:02
• @DietrichBurde The point is that "the left-inverse in the group $O(n)$ is the right-inverse" is an equivalent statement to what I'm asking a proof for. So, this does not help. May 29, 2020 at 13:21
• But this holds in any group. This is another duplicate and has been proved several times here. See for example here. May 29, 2020 at 13:24

Counterexample for $$m \neq n$$:

$$A = \begin{bmatrix} 1/\sqrt{2} \\ 1/\sqrt{2} \end{bmatrix} \in \mathbb{R}^{2 \times 1}$$

For $$m = n$$ the statement is true because then $$A^T = A^{-1}$$, as inverses are unique.

• Please take note of my edit. May 29, 2020 at 14:37

If $$m = n$$, the statement is true, because any left inverse is also a right inverse and vice versa. This fact is proved here: If $AB = I$ then $BA = I$

If $$m \neq n$$, the statement is false. A counterexample is provided by the matrix $$A = \begin{bmatrix} 1 & 0 \\ 0 & 1 \\ 0 & 0\end{bmatrix}.$$ For another counterexample we could take $$A = \begin{bmatrix} 1 \\ 0 \end{bmatrix}.$$

If $$m > n$$, then the rank of $$A A^T$$, which we know is equal to the rank of $$A^T$$, is at most $$n$$. So $$A A^T$$ does not have full rank, so it is not the identity matrix.

• Can you provide a proof? May 29, 2020 at 10:59
• The first statement, that for a square matrix any left inverse is also a right inverse and vice versa, is proved here: math.stackexchange.com/questions/3852/if-ab-i-then-ba-i May 29, 2020 at 11:00
• You should give a counterexample for $n \ne m$. May 29, 2020 at 11:01
• Please take note of my edit. May 29, 2020 at 14:36

Yes. This is because orthonormal columns of matrix implies orthonormal rows of matrix. Check here.

This assumes $$m=n$$, otherwise, take $$A =\begin{bmatrix}1 & 0 \\0&1\\0&0\end{bmatrix}$$

We get $$A'A=I$$, but $$AA'\neq I$$

• It seems like the answer in your link is concluding $(1)$ implies that $A^T$ is the actual inverse (not only the left-inverse) of $A$, but this implication is essentially a restatement of the question I'm asking here. May 29, 2020 at 11:00
• Please take note of my edit. May 29, 2020 at 14:36