Insight on the polar decomposition of a shear?

Question

I recently learned it, and really love the polar decomposition of a matrix, because it was the first time I actually could picture what it meant to "apply a transformation to space" (a phrase I kept on seeing being thrown about in forums and videos).

We can picture any linear transformation as scaling along orthogonal directions (along the eigenvectors of $S$ in the decomposition), and then rotating. $$A=QS$$

Decompositions have helped me understand lots of properties about linear transformations. What I'm wondering right now is if there's any special or enlightening characteristics about the polar decomposition of a shear.

I'm not sure if "shear" is the correct term, but I mean a transformation like this one:

\begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}

This transformation doesn't rotate space, and it only has one eigenvector. I tried using a polar decomposition calculator online, but it kept on failing me...

Any help? Any insights? Thanks!

Answer 1

This post has been triggered by mentioning 'ugly' in the comments ...

G. H. Hardy once wrote (in "A Mathematician's Apology", 1941)
Beauty is the first test: There is no permanent place in the world for ugly mathematics.
See also on Philosophy.SE the post What did Hardy mean by “ugly mathematics”? .

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Let software work on $A=\left(\begin{smallmatrix} 1&1\\0&1\end{smallmatrix}\right)$

python.scipy.linalg.polar(A) yields the polar decomposition

 Q = [[ 0.89442719,  0.4472136 ],
      [-0.4472136 ,  0.89442719]]

 S = [[ 0.89442719,  0.4472136 ],
      [ 0.4472136 ,  1.34164079]]

Notice that $\,0.4472136=1\big/{\sqrt 5}\,$.

Manual handling of $A$

We determine $\,S=|A|\,$ first. $$S^2\,=\,A^*\!A\:=\:\begin{pmatrix}1&1\\1&2\end{pmatrix} \;\text{ has the eigenvalues } \left(\frac{\sqrt 5\pm 1}2\right)^2\,,$$ hence it's positive-definite.
$\frac{\sqrt 5+1}2=1.618034$ is the Golden ratio, which in the sequel is denoted by $\phi$. Note that it satisfies

• $\phi^2=\phi +1\iff \phi(\phi -1) =1$
• $\phi +2=\sqrt5\,\phi$

This helps in identifying eigenvectors of $S^2$ as $$\begin{pmatrix}1&1\\1&2\end{pmatrix} \begin{pmatrix}1\\ \phi\end{pmatrix} \:=\: \phi^2\begin{pmatrix}1\\ \phi\end{pmatrix}\quad\text{and}\quad \begin{pmatrix}1&1\\1&2\end{pmatrix} \begin{pmatrix}-\phi\\ 1\end{pmatrix} \:=\: (\phi-1)^2\begin{pmatrix}-\phi\\1\end{pmatrix}\,.$$ They are also eigenvectors of $S$ to the eigenvalues $\phi$ and $\phi -1$.
Incorporating normalisation factors one gets $$\begin{align} S \:& =\frac1{\sqrt{\phi^2+1}}\begin{pmatrix}1&-\phi\\ \phi&1\end{pmatrix} \begin{pmatrix}\phi&0\\0&\phi-1\end{pmatrix} \frac1{\sqrt{\phi^2+1}}\begin{pmatrix}1&\phi\\ -\phi&1\end{pmatrix} \:=\: \frac1{\phi+2}\begin{pmatrix}2\phi&\phi\\ \phi&3\phi\end{pmatrix} \\[2ex] & =\frac1{\sqrt 5}\begin{pmatrix}2&1\\1&3\end{pmatrix} \end{align}$$

As $S\,$ is positive-definite, thus invertible, the unitary factor $Q$ in the polar decomposition may be obtained as $$Q\,=\,AS^{-1}\:=\:\begin{pmatrix}1&1\\0&1\end{pmatrix}\,\frac1{\sqrt 5}\begin{pmatrix}3&-1\\-1&2\end{pmatrix}\;=\; \frac1{\sqrt 5}\begin{pmatrix}2&1\\-1&2\end{pmatrix}$$ In summary $$A\:=\:\frac1{\sqrt 5}\begin{pmatrix}2&1\\-1&2\end{pmatrix}\: \frac1{\sqrt 5}\begin{pmatrix}2&1\\1&3\end{pmatrix}\,,$$ and where's the ugliness?