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Given an nxn matrix A with mutually orthonormal columns, how does one show that A has mutually orthonormal rows WITHOUT assuming that A is orthogonal (A^T = A^-1)? I can show that A has mutually orthogonal rows by using the orthogonality between the row space and null space of a matrix, as well as the fact that the columns of A span R^n, but I need to show that the rows also have unit length. That is where I am getting stuck.

A similar question was asked here: Orthonormal columns and rows ; however, in trying to prove that A is orthogonal iff A has orthonormal columns, they don't show that AA^T = I given orthonormal columns, so their proof seems incomplete (if A is orthogonal, then A^T *A = I AND AA^T = I). To finish the proof, I believe they would need to answer my question.

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  • $\begingroup$ Usually by this point one has already established that if $AB=I$ then $BA=I$. $\endgroup$ – user856 Jan 19 '20 at 6:01
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The dot product of the $i$'th and $j$'th columns of $A$ is $(A^T A)_{ij}$, so $A^T A = I$ is equivalent to the statement that the columns are orthonormal. Now this implies that $A$ has rank $n$, and therefore is invertible, and therefore $A^T = A^{-1}$, so $A A^T = I$ also, which is equivalent to the statement that the rows are orthonormal.

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The canonical answer is Robert's. Here's another argument:

You know that $A^TA=I$. Multiplying by $A$ on the left and $A^T$ on the right, you get $(AA^T)^2=AA^T$. So $AA^T$ is an idempotent. In particular its eigenvalues can only be $0$ and $1$. Let $m$ be the multiplicity of $1$ as an eigenvalue of $AA^T$. Then $$ m=\operatorname{Tr}(AA^T)=\operatorname{Tr}(A^TA)=\operatorname{Tr}(I)=n. $$ This shows that all eigenvalues of $AA^T$ are $1$. Thus it is invertible; as the only invertible idempotent is $I$ we get that $AA^T=I$.

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