Given two invertible linear transformations T1,T2 in L(V) that preserve angles i.e. $\frac{(T(u), T(v))} {∥T(u)∥∥T(v)∥} = \frac{(u, v)} {∥u∥∥v∥} $. How can I show that T1T2 and T-1 also preserve angles?
From the definition I believe T1,T2 to be isometries but I'm not sure if this helps or is true.
Thanks for you help!
Note that
$$\frac{\langle(T_2\circ T_1)(u), (T_2\circ T_1)(v)\rangle} {∥(T_2\circ T_1)(u)∥∥(T_2\circ T_1)(v)∥} =\frac{\langle T_1(u), T_1(v)\rangle} {∥T_1(u)∥∥T_1(v)∥}= \frac{\langle u, v\rangle } {∥u∥∥v∥}$$
where in the first equality we have used that $T_2$ preserves angles and in the second equality that $T_1$ preserves angles. Thus $T_2\circ T_1$ preserves angles.
Now, assume $T^{-1}$ exists. Then
$$\frac{\langle u, v\rangle } {∥u∥∥v∥}=\frac{\langle(T\circ T^{-1})(u), (T\circ T^{-1})(v)\rangle} {∥(T\circ T^{-1})(u)∥∥(T\circ T^{-1})(v)∥} =\frac{\langle T^{-1}(u), T^{-1}(v)\rangle} {∥T^{-1}(u)∥∥T^{-1}1(v)∥},$$ which shows that $T^{-1}$ preserves angles.
Finally, note that $T(x)=cx$ preserves angles for any $c\ne 0.$ However it is not an isometry unless $c=\pm 1.$