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).

  • $\begingroup$ A matrix is orthogonal if and only if its columns form an orthonormal basis, which is the same as saying that $Q^T=Q^{-1}$. Let $R=Q^T$. Then $R^T=Q=(Q^T)^{-1}=R^{-1}$. So $R$ is orthogonal. Where is the problem? $\endgroup$ – egreg Mar 24 '16 at 17:11
  • $\begingroup$ @egreg the problem is that you implicitly use the two-sidedness of matrix inverses, which (I find, anyway) is not intuition-friendly. Consider, for example, why it is possible for a matrix to have orthogonal columns without having orthogonal rows. The key here is to make that two-sidedness of the inverse understandable, apparently. $\endgroup$ – Omnomnomnom Mar 24 '16 at 17:14
  • $\begingroup$ @egreg I don't have a problem understanding why both the rows as well as the columns are orthonormal mathematically. but GIlbert Strang in his book (check out the extract I posted) has said he couldn't see geometrically why the rows are forced to be orthonormal given the columns were. I wanted to just confirm if my geometric explanation had any fallacies. $\endgroup$ – Aditya Mar 24 '16 at 17:18
  • $\begingroup$ @Shikamaru you have the right idea. Namely: with linear maps in finite-dimensional spaces, if $AB$ is the identity map, then $BA$ must also be the identity map. Of course, then, this is true for orthogonal maps, which happen to be rotations/reflections. $\endgroup$ – Omnomnomnom Mar 24 '16 at 17:19
  • $\begingroup$ @Shikamaru I would say, however, that your explanation is not a "geometric" one in the sense that Strang is going for. That is, if we have a set of $n$ row-vectors in $n$-dimensional space, why should the resulting columns be orthonormal as well? To see this, we would want a geometric way to go directly from $n$ row-arrows to $n$ column-arrows. $\endgroup$ – Omnomnomnom Mar 24 '16 at 17:23

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