Inequality from linear algebra I will be greateful for some hints in proving the following inequality from linear algebra:
Let $L:\mathbb{R}^n\to\mathbb{R}^n$ be a linear inveriable map and let $u$, $v$ be nonzero vectors. Prove that $$\frac{\sin \mathrm{angle}\,(Lu,Lv)}{\sin\mathrm{angle}\,(u,v)} \le \|L\|\|L^{-1}\|,$$ which is translated as
$$\frac{\|Lu\|^2\|Lv\|^2 - (Lu,Lv)^2}{\|Lu\|^2\|Lv\|^2} \cdot \frac{\|u\|^2\|v\|^2}{\|u\|^2\|v\|^2 - (u,v)^2} \le \|L\|^2\|L^{-1}\|^2.$$
 A: Presumably $u$ and $v$ are linearly independent and you are talking about the induced 2-norm. Denote the span of $u$ and $v$ by $S$. Since the induced 2-norm is orthogonally invariant and the (unsigned) angle between two vectors is preserved under orthogonal transform, you may assume that $Lu,Lv\in S=\mathbb R^2$ (here $\mathbb R^2$ means the vector subspace spanned by the first two vectors in the standard basis of $\mathbb R^n$). Let the restriction of $L$ on $S$ be $M$. Then $\|L\|\ge\|M\|$ and $\|L^{-1}\|\ge\|M^{-1}\|$. So, it suffices to show that
$$
\frac{\left|\sin \mathrm{angle}\,(Mu,Mv)\right|}
{\left|\sin\mathrm{angle}\,(u,v)\right|}
\le \|M\|\|M^{-1}\|.\tag{1}
$$
Since $\|x\times y\|=\|x\|\|y\|\left|\sin \mathrm{angle}\,(x,y)\right|$, if we let $Mu=au+bv$ and $Mv=cu+dv$, then $(1)$ can be rewritten as
$$
|ad-bc|\frac{\|u\|\|v\|}{\|Mu\|\|Mv\|}\le \|M\|\|M^{-1}\|.\tag{2}
$$
Note that $ad-bc=\det(M)$ and $\frac{\|u\|\|v\|}{\|Mu\|\|Mv\|}\le\|M^{-1}\|^2$. Now the inequality $(2)$ is obvious if you express $|\det(M)|,\|M\|$ and $\|M^{-1}\|$ in terms of the two singular values of $M$.
