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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?

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    $\begingroup$ Does this answer your question? Does $AA^T=I$ imply that $A^TA=I$? - and conversely. $\endgroup$
    – Dietrich Burde
    May 29, 2020 at 10:59
  • $\begingroup$ @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). $\endgroup$
    – 0xbadf00d
    May 29, 2020 at 11:02
  • $\begingroup$ 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. $\endgroup$
    – Dietrich Burde
    May 29, 2020 at 11:02
  • $\begingroup$ @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. $\endgroup$
    – 0xbadf00d
    May 29, 2020 at 13:21
  • $\begingroup$ But this holds in any group. This is another duplicate and has been proved several times here. See for example here. $\endgroup$
    – Dietrich Burde
    May 29, 2020 at 13:24

3 Answers 3

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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.

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  • $\begingroup$ Please take note of my edit. $\endgroup$
    – 0xbadf00d
    May 29, 2020 at 14:37
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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.

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  • $\begingroup$ Can you provide a proof? $\endgroup$
    – 0xbadf00d
    May 29, 2020 at 10:59
  • $\begingroup$ 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 $\endgroup$
    – littleO
    May 29, 2020 at 11:00
  • $\begingroup$ You should give a counterexample for $n \ne m$. $\endgroup$
    – Paul Frost
    May 29, 2020 at 11:01
  • $\begingroup$ Please take note of my edit. $\endgroup$
    – 0xbadf00d
    May 29, 2020 at 14:36
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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 $AA'\neq I$

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  • $\begingroup$ 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. $\endgroup$
    – 0xbadf00d
    May 29, 2020 at 11:00
  • $\begingroup$ Please take note of my edit. $\endgroup$
    – 0xbadf00d
    May 29, 2020 at 14:36

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