I just want to know if my solutions are correct for the following problem (euclidean norm assumed):
A linear transformation $T:\Bbb R^n \to \Bbb R^n$ is norm preserving if $|T(x)| = |x|$ for all $x \in \Bbb R^n$, and inner product preserving if $\left\langle T(x),T(y)\right\rangle = \left\langle x, y\right\rangle$.
$(a)$ Prove that $T$ is norm preserving if and only if $T$ is inner product preserving.
<p>$(b)$ Prove that such a linear transformation is $1$-$1$ and $T^{-1}$ is of the same sort.</p>
My solution for $(a)$ was: suppose $T$ is inner product preserving, then since $|x|=\sqrt{\left\langle x,x \right\rangle}$ we have that $|T(x)| = \sqrt{\left\langle T(x),T(x)\right\rangle}=\sqrt{\left\langle x, x\right\rangle} = |x|$ so that $T$ is norm preserving. Suppose now that $T$ is norm preserving, so we have the polarization identity:
$$\left\langle T(x), T(y)\right\rangle=\frac{|T(x)+T(y)|^2 - |T(x)-T(y)|^2}{4}$$
But by virtue of linearity $T(x)+T(y) = T(x+y)$ and $T(x)-T(y)=T(x-y)$ and so since $T$ preserves norms we have $\left\langle T(x),T(y)\right\rangle = \left\langle x,y\right\rangle$ so that $T$ is inner product preserving also.
My solution for $(b)$ was: Suppose that $T(x) = 0$, then we have $|T(x)| = 0$, but since $T$ is norm preserving $|x| = 0$ and this implies $x = 0$ so that $T$ is injective. Also by the rank-nullity theorem $T$ is surjective because $\dim \ker T = 0$ and so $\dim \operatorname{Im} T = \dim \Bbb R^n$. Now, because of that, given $y \in \Bbb R^n$ there's some $x \in \Bbb R^n$ such that $ y = T(x)$. In that case, consider the norm of $T^{-1}(y)$, because of the analysis that there's some $x$ such that $y= T(x)$ we will have:
$$T^{-1}(y) = x$$
Now recalling that $T$ preserves norms, we have that $|x| = |T(x)|$, but $T(x) = y$ so that $|x| = |y|$, so taking the norm on the above equation we finally have:
$$\left|T^{-1}(y)\right| = |x| = |T(x)| = |y|$$
Are my solutions correct? Thanks very much in advance!
