Unitary and Orthogonal Operators 
Let $T$ be a linar operator on $V$ over $F$, a finite-dimensional inner product space. If $\|T(x)\| = \|x\|$, and $F$ is $\mathbb{C}$, We say that $T$ is unitary. Similarly, if $F$ is $\mathbb{R}$, we say that $T$ is orthogonal.

Outside of this definition, what is the intuition behind unitary/orthogonal operators? It seems the length of $T(x)$ is the same as the length of $x$; does that mean $T$ has no real effect on $x$ whatsoever? What exactly is going on with this operator? Thank you.
 A: If $T$ is orthogonal, then $\langle x,y\rangle = \langle Tx,Ty\rangle$ for every pair of vectors $x,y$.
Notice that that happens, if, for example, $T$ rotates every vector $43^\circ$ about a specified axis.
In the plane, the mappling $\begin{bmatrix} x \\ y \end{bmatrix} \mapsto \begin{bmatrix} \cos43^\circ & -\sin43^\circ \\ \sin 43^\circ & \phantom{+}\cos 43^\circ \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix}$ just rotates the input vector $43^\circ$ counterclockwise about the origin. In matrix notation, suppose $A$ is an $n\times n$ orthogonal matrix. Then $\langle Ax,Ay\rangle = (Ax)^T(Ay) = x^T A^T A y ,$ and this must be equal to $x^T y$ for every pair of vectors $x,y$.  In particular, that implies it must be true if $x,y$ are any pair of standard basis vectors, and that implies $A^TA = I_n,$ the identity matrix. Thus a matrix is orthogonal precisely if its transpose is its inverse. That has this consequence:
$$
1 = \det I_n = \det(A^T A) = (\det A^T)(\det A) = (\det A)^2,
$$
so it follows that
$$
\det A = \pm1.
$$
The set of orthogonal matrices for which the determinant is $+1$ corresponds to the set of rotations.  In $3$-dimensional space, that means rotation about an axis that is a straight line. Which line it is, and through how many degrees does it rotate, determines which $3\times3$ orthogonal matrix it is. Then ones whose determinant is $-1$ reverse the orientation, so that in $2$-dimensional space, they transfrom the letter R to a backward R, and in $3$-dimensional space they transform a left shoe to a right shoe.
With complex numbers one uses conjugate-transposes rather than just transposes, and the determinant, rather than being either $+1$ or $-1,$ is some complex number whose absolute value is $1,$ so it's on the circle of unit radius centered at $0.$
A: Let's consider real inner product spaces first. The length between two points $\mathbf{x},\mathbf{y}$ is given by taking the norm of the displacement, $\|\mathbf{x}-\mathbf{y}\|$. If a map is linear, then it preserves the distance between points if and only if it preserves the norm of all elements. Thus, it is an isometry. Not only that, but since we can define $\langle x,y\rangle $ in terms of $\|\cdot\|$ using the polarization identity,
$$\langle x,y\rangle=\frac{1}{2}\left(\|x+y\|^2-\|x\|^2-\|y\|^2\right),$$
and since $\langle x,y\rangle=\|x\|\|y\|\cos\theta$, these isometries also preserve the angles between vectors.
In $\mathbb{R}^3$ these include the rotations and reflections of space. A reflection is where we take a two-dimensional plane through the origin and flip all vectors across it. In higher dimensions the same thing is true. The group of rotations in higher dimensions is generated by rotations of two-dimensional subspaces, reflections are across hyperplanes (which have one dimension less than the whole space) and every isometry is a composition of rotations and reflections.
For complex inner product spaces we simply mimic the definition for real inner product spaces in order to get a new thingie to study.
