4
$\begingroup$

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.

$\endgroup$
0
3
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.