Showing $\Vert H \Vert_2 = \Vert MN \Vert_2 \le \Vert M \Vert_2 \Vert N \Vert_2 = \Vert N \Vert_2$ is actually a strict equality?

Question

1. If I have matrices $H$, $M$ and $N$ such that $H = MN$ and $M$ is orthogonal then I have

$$\Vert H \Vert_2 = \Vert MN \Vert_2 = \Vert N \Vert_2,$$ as orthogonal matrices are invariant under the 2-norm.

2. However if I do the following I only get a inequality instead of a strict equality. $M$ invariant under the 2-norm implies that $\Vert M \Vert_2 = 1$. Then

$$\Vert H \Vert_2 = \Vert MN \Vert_2 \le \Vert M \Vert_2 \Vert N \Vert_2 = \Vert N \Vert_2$$

Why do these two results conflict with each other? Shouldn't it be possible to prove a strict inequality in 2nd case?

Answer 1

They do not conflict each other as in the first item, you used the fact that $M$ is orthogonal and deduced that $||MN||_2 = ||N||_2$ while in the second item, you used the (weaker) fact that $||M||_2 = 1$ to deduce that $||MN||_2 \leq ||N||_2$. It is a good exercise to construct a non-orthogonal $M$ with $||M||_2 = 1$ such that $||MN||_2 < ||N||_2$ (try to take $M$ to be a matrix whose all entries are identical).