Identifying a “rotated shear” matrix

linear algebralinear-transformationsmatricesmatrix decompositionorthogonal matrices

Suppose that we're in $\Bbb R^n$. Then the simplest shear matrix can be described as

$$S_\lambda =
\begin{bmatrix}
1 & \lambda & 0 & \ldots & 0 \\
0 & 1 & 0 & \ldots & 0 \\
0 & 0 & 1 & \ldots & 0 \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
0 & 0 & 0 & \ldots & 1
\end{bmatrix}
$$

Where $\lambda$ is a free parameter that can be thought of as the "slope" of the shear.

This shear preserves the $(n-1)$-dimensional subspace that is orthogonal to the vector $(0,1,0,…,0)$, and shears everything along the $(1,0,0,…,0)$-axis. I will call this the "canonical" shear basis.

If we would like to shear along a different axis and preserve a different $(n-1)$-dimensional subspace, that can be expressed via the following matrix product:

$$
OS_\lambda O^{-1}
$$

where $O$ is an arbitrary orthogonal matrix (which may have determinant $-1$). This can be thought of as rotating and/or reflecting our desired vector and subspace into the canonical basis, performing the shear, and then rotating back. I will call this a "rotated shear."

My question: given a "rotated shear" matrix, how can I identify it as such, and decompose it into the product $OS_\lambda O^{-1}$? Is there some easy way to identify the rotation, and the shear axes involved?

Or, put another way, is there a general matrix decomposition that will yield the desired form if I give it a rotated shear matrix (and will yield something else for any other general matrix)?

Best Answer

An $\ n\times n\ $ matrix $\ A\ $ is of the form $\ OS_\lambda O^{-1}\ $, with $\ O\ $ orthogonal and $\ \lambda\ne 0\ $, if and only if:

  • The nullspace $\ \mathcal{N}_{A-I}\ $ of $\ A-I\ $ has dimension $\ n-1\ $; and
  • $\ \left(A-I\right)^{\,2} = 0\ $.

For if $\ A=OS_\lambda O^{-1} $, then $\ \mathcal{N}_{A-I}\ $ is the space spanned by the first and third to $\ n^\mathrm{\,th}\ $ columns of $\ O\ $, and $$\ \left(A-I\right)^{\,2} =O\left(S_\lambda -I\right)^2 O^{-1} =0\ .$$

Now suppose that $\ A\ $ satisfies the above two conditions. Then there exists a non-zero $\ v\in \mathcal{N}_{A-I}^\top\ $, and since $\ v\not\in \mathcal{N}_{A-I}\ $ then $\ \left(A-I\right)v\ne 0\ $. But since $\ \left(A-I\right)^{\,2} = 0\ $ then $\ \left(A-I\right)v\in \mathcal{N}_{A-I}\ $, and so $\ Av = v + \omega\ $ for some non-zero $\ \omega\in \mathcal{N}_{A-I}\ $. Thus, $\ A\frac{v}{||v||}=\frac{v}{||v||}+\left(\frac{||w||}{||v||}\right)\frac{w}{||w||}\ $. Now if $\ \omega_1,\omega_2, \dots, \omega_n\ $ is any orthonormal basis with $\ \omega_1= \frac{w}{||w||}\ $ and $\omega_2= \frac{v}{||v||}\ $, and $\ \Omega\ $ the matrix whose $\ j^{\,\mathrm{th}}\ $ column is $\ \omega_j\ $, then $\ \Omega^{-1}A\Omega=S_\lambda\ $, where $\ \lambda = \frac{||w||}{||v||}\ $.

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