For an orthogonal matrix $Q$, prove $\operatorname{cond}(Q)=1$

Question

Given an orthogonal matrix $Q$, prove $$\|Q\|_2\cdot \|Q^{-1}\|_2=1$$

I succeed to solve it with eigenvalues but I'm looking for an easier way.

Answer 1

An orthogonal matrix preserves distance (it's an isometry). It means that for any $\mathbf x$ we have $\|Q\mathbf x\|_2=\|\mathbf x\|_2$.

So: $$\|Q\|_2 \overset{\text{def}}= \sup_{\mathbf x\ne \mathbf 0}\frac{\|Q\mathbf x\|_2}{\|\mathbf x\|_2} = \sup_{\mathbf x\ne \mathbf 0}\frac{\|\mathbf x\|_2}{\|\mathbf x\|_2}=1$$ Since $Q^{-1}$ is orthogonal as well, the result follows.