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

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

I came across this when studying linear transformations in these notes which says:

The similarity group is obtained from the affine group by requiring that $$A$$ be orthogonal: $$A^TA=\lambda^2I$$

Can't seem to wrap my head around this one.

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.

Orthogonality is forced by requiring a scalar multiple of the identity; if any two columns $$a_i, a_j$$ were not orthogonal, there would be a nonzero entry in the corresponding entries $$b_{ij}, b_{ji}$$ of the product.

Orthogonality results in the absence of any shear/twist in the transformation, restricting it to only reflection, rotation and translation.

The eigenvalue squares as scalars allow for non-normality, as it is possible to scale the original space without affecting the angles between lines in it, provided all axes of the space are scaled by the same amount. If $$A$$ was instead orthonormal, you force a lack of scaling as well.

See also this question, for which answers explain a converse point, why similarity transformation is a subtype of affine transformation.

Let $$A=RU$$ be the polar decomposition of $$A$$, where $$R$$ is positive semi-definite and $$U$$ is unitary.

Then $$R^2=A^TA=\lambda^2I$$; being diagonalizable this forces $$R=|\lambda|I$$ and thus $$A=|\lambda|U$$, a similarity transformation.