Why does an affine transformation $A$ when constrained by $A^TA=\lambda^2I$ result in a similarity transformation?

Let $\mathbf {x, y} \in \mathbb{R}^n$ be unit vectors.
Then, the angle $\theta_1$ between them is given by $ \cos \theta_1 = \mathbf{x^\top y}$.

Also, angle between $\mathbf{Ax}$ and $\mathbf{Ay}$ is given by :

$\cos \theta_2 = \dfrac{(\mathbf{Ax})^\top (\mathbf{Ay})}{\|\mathbf{Ax}\|\|\mathbf{Ay}\|}$.
Now, $\|\mathbf{Ax}\|_2^2 = (\mathbf{Ax})^\top(\mathbf{Ax}) = \mathbf{x^\top A^\top Ax} = \mathbf{\lambda ^2 x^\top x}$.
Thus, $\mathbf{\|Ax\|} = \mathbf{|\lambda|}$.
$\implies \cos \theta_2 = \mathbf{\dfrac{\lambda^2 x^ \top y}{\lambda^2}} = \mathbf{x^ \top y} \implies \theta _1 = \theta _2$.
Thus, $\mathbf{A}$ preserves angles and is a similarity transformation.

Tags:

Linear Algebra

Matrices

Linear Transformations

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