How do we have ${\mathbf {v} }^{\mathrm {T} }{\mathbf {v} }={\mathbf {v} }^{\mathrm {T} }Q^{\mathrm {T} }Q{\mathbf {v} }$?

Question

From Wikipedia, if a linear transformation, in matrix form $Q{\mathbf {v}}$, preserves vector lengths, then:

$${\displaystyle {\mathbf {v} }^{\mathrm {T} }{\mathbf {v} }=(Q{\mathbf {v} })^{\mathrm {T} }(Q{\mathbf {v} })={\mathbf {v} }^{\mathrm {T} }Q^{\mathrm {T} }Q{\mathbf {v} }.}$$

What happens to $Q^{\mathrm {T} }Q$?

Answer 1

If $Q$ preserves vector lengths, then $Q$ is orthogonal, thus $Q^TQ=I.$

Answer 2

To add some examples to Fred's answer you could for take matrices that realize reflections or rotations.

If one exchanges the rows of the identity matrix you get a transformation that preserves length: $Q_1=\begin{array}{rrrr} 0 & 0 & 0 & 1 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 1 & 0 & 0 & 0 \\ \end{array}$

You can check that $Q_1 {Q_1}^T=I$ and for any vector $v\in\mathbb{R}^4$ one gets $||Qv||=||v||$.

Another example is a 2D rotation with angle $\phi$ that is realized by the matrix

$Q_2=\begin{array}{rrrr} cos \phi & -sin \phi \\ sin \phi & cos \phi \\ \end{array}$