Orthornomal matrices Is there a more direct reason for the following:

If the columns of $n\times n$ square matrix are orthonormal, then its rows are also orthonormal.

The standard proof involves showing that left inverse of a matrix is same as the right inverse and thereby concluding that if $Q^TQ = I$, then $QQ^T = I$. This seems to be more of an algebraic manipulation. Can someone offer me a geometric insight?

Thanks

If $Q$ is orthonormal, its transpose is its inverse, and vice versa. Understanding this is key to your question. You are asking for a geometric understanding of this, rather than a symbolic algebra understanding. So let's give that a try.
Because columns are orthonormal, then what does $Q^T$ do to the $i$th column of $Q$? If you are in the habit of seeing matrix multiplication with a vector as a sequence of dot products of rows with that vector (which captures the degree to which the vector is parallel to each of the rows), then you see that $Q^T$ takes the $i$th column to the $i$th standard basis vector.
But then what does $Q$ do to the $i$th standard basis vector? Again if you see matrix multiplication with a vector as a sequence of dot products, then you see that $Q$ applied to the $i$th standard basis vector captures the projection of each of $Q$'s rows onto the $i$th standard basis vector, yielding $Q$'s $i$th column.
We've just established that $Q$ and $Q^T$, as actions on vectors, exchange the set of columns of $Q$ with the set of standard basis vectors. So they are inverse actions.
