# Given an $n \times n$ matrix $A$ with orthonormal columns, how does one show that $A$ has orthonormal rows?

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.

• Usually by this point one has already established that if $AB=I$ then $BA=I$. – user856 Jan 19 '20 at 6:01

## 2 Answers

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.

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 $$AA^T=I$$.