Norm inequality for linear isomorphism I want to show that if $L:\mathbb{R}^d\rightarrow\mathbb{R}^d$ is a linear isomorphism and $V_1,V_2$ are linear subspaces of $\mathbb{R}^d$ then:
$$\frac{1}{||L||\space||L^{-1}||}\leq\frac{|sin\measuredangle(L(V_1),L(V_2))|}{|sin\measuredangle(V_1,V_2)|}\leq||L||\space||L^{-1}||$$
where the angle $\measuredangle (V,W)$ between two subspaces $V$ and $W$ of $\mathbb {R}^d $ is
the smallest angle between non-zero vectors $v\in V$ and $w\in W$.
I tried to solve the first inequality using the relation between the sinus and the vector product but I think that it is not the correct way.
Can someone help me? Thank you.
 A: First notice that one inequality implies the other by applying $L^{-1}$ instead of $L$, so we will prove only the right one.
Secondly, if the angle between two given vectors $u \in V_1$ and $v \in V_2$ is greater than $\frac{\pi}{2}$, then the angle between $-u$ and $v$ is less than $\frac{\pi}{2}$, so there is no loss of generality by assuming that all vectors we consider have angle less than $\frac{\pi}{2}$. The angle between $u$ and $v$ is the same as the angle between $\frac{u}{|u|}$ and $\frac{v}{|v|}$, so we will also only consider unit vectors. Now take sequences $u_n \in V_1$ and $v_n \in V_2$ such that the angle between $u_n$ and $v_n$ converge to the angle between $V_1$ and $V_2$, since the first sequence is contained in $\mathbb{S}^{d-1} \cap V_1$ and the second one in $\mathbb{S}^{d-1} \cap V_2$, both of these sets being compact sets, we may take subsequences if necessary and suppose that $\lim u_n = u$ and $\lim v_n = v$, that is, the smallest angle is obtained by a pair of vectors, we will show the inequality for this pair of vectors.
Also, composing by an isometry if necessary, we may suppose without loss of generality that $Lu$ is colinear with $u$ and the subspace $V$ generated by $u,v$ is invariant by $L$. Consider an othornomal basis in $V$ where $u$ is the first element of that basis. In such basis, the matrix associated with $L$ in V is written as

\begin{pmatrix}
  a & b \\
  0 & d
 \end{pmatrix}

If $v$ is orthogonal to $u$ then it's obvious since $\|L\|\|L^{-1}\| \geq 1$, therefore, writing $u = (1,0)$ we take a representative of $v$ in the form $v = (1,y)$ which implies $Av = (a+yb,yd)$ and
$$
 \frac{\sin\angle(LV_1,LV_2)}{\sin\angle(V_1,V_2)} \leq \frac{\sin(\min(\angle(Lu,Lv),\angle(Lu,-Lv))}{\sin\angle(u,v)} = 
 \frac{|y||d|}{|Lv|}\frac{|v|}{|y|} =\\= \left|L\left(\frac{v}{|v|}\right)\right|^{-1}|d| \leq \|L^{-1}\| \|L\|
 $$
