I was reading the book - Linear Algebra and its Applications, when I saw -
Remark 2. Since $Q^T = Q^{-1}$, we also have $QQ^T = I$. When Q comes before $Q^T$, multiplication takes the inner products of the rows of Q. (For $Q^TQ$ it was the columns.) Since the result is again the identity matrix, we come to a surprising conclusion: The rows of a square matrix are orthonormal whenever the columns are. The rows point in completely different directions from the columns, and I don’t see geometrically why they are forced to be orthonormal—but they are.
I was wondering if the following geometric interpretation is valid.
As we already know geometrically, an orthogonal matrix Q is the product of a rotation (through say θ) and a reflection. When Q is a square matrix, then $Q^T$ ( Q transpose) will be $Q^{-1}$ (Q inverse). And therefore Q^T ( Q transpose) can be thought of geometrically as the undoing of Q, i.e. rotation about -θ and a repeated reflection about the same axis. Thus operating $QQ^T$ is same as operating $Q^TQ$, as we are doing (rotation reflection) then (opposite rotation and reflection) - $QQ^T$ or (opposite rotation and reflection) and then (rotation reflection) - $Q^TQ$,
Both of which do nothing to the vector (being operated on). ie they are the identity matrix.
From the above as we see that $Q^T$ ( Q transpose) can be thought of geometrically as the undoing of Q, a product of a rotation and a reflection. we can say that the rows are also orthonormal (as they form the columns of Q^T).