Apparently the following matrix $$A=\begin{bmatrix} 2 & 4 \\\ 1 & 2 \end{bmatrix}$$ is a composition of two linear transformations, $BC=A$. The objective of this exercise is to decompose $A$ into $B,C\neq I_2$ such that $B$ and $C$ are either a rotation, scaling, projection, shear, or reflection.

It seems that all points end up on the line $y=\frac{1}{2}x$, so I tried making $C$ a projection onto the vector $\langle 2,1\rangle$, but then I cannot find a transformation $B$ that works.

  • $\begingroup$ Any linear transformation is a composition of two linear transformations. You can take $B=A,C=I$. $\endgroup$ – Kavi Rama Murthy Apr 18 at 7:36
  • $\begingroup$ @KaviRamaMurthy Yes, that is certainly true but my exercise restricts me from using $C=I$. I apologize for not clarifying. $\endgroup$ – 高田航 Apr 18 at 7:44
  • $\begingroup$ Then take $B=2A$, $C=\frac 1 2 I$. $\endgroup$ – Kavi Rama Murthy Apr 18 at 7:45
  • $\begingroup$ I edited my question to hopefully provide more clarification. I need for $B$ and $C$ to carry geometric significance in the sense that it applies a certain rigid motion to a vector. $\endgroup$ – 高田航 Apr 18 at 7:47

We want $BC=A$ such that $B$ and $C$ represent simple transformations (scaling etc).

As noted, all the points end up on the line $y=\frac 12 x$. Therefore either $B$ or $C$ must be a projection.

Using $A$ we see that point $(1,0)$ transforms to point $(2,1)$, and $(0,1)$ to $(4,2)$. If we sketch these four points on a graph we see that the only realistic option for the other transformation is scaling. Moreover, the geometry suggests that scaling must happen first. Therefore $B$ is projection and $C$ is scaling.

Given that projection onto $y=\frac 12 x$ is orthogonal, point $(2,1)$ can be projected back to the $x$-axis (since it originated from $(1,0)$) perpendicular to $y=\frac 12 x$. And by looking at similar triangles, the $x$ scale factor can be found to be $\frac 52$.

The $y$ scale factor can be found similarly by projecting $(4,2)$ back to the $y$-axis.

From this $C$ can be assembled and then its inverse used to solve the matrix equation for $B$ (the projection).


Answer to the question before it was edited: for any non-singular matrix $S$ we can write $A= (AS) S^{-1}$ and we can take $B=AS, C=S^{-1}$.


You can see that $A$ has rank one, since column 2 is 2 times column 1. If the only operations allowed are

  1. rotation
  2. scaling
  3. projection
  4. shear
  5. reflection

then the only one of these which reduces rank is a projection. So there must be at least one projection.

For symmetry reasons we see that we can divide $A$ by $\begin{bmatrix}0&1\\1&0\end{bmatrix}$ to get $\begin{bmatrix}4&2\\2&1\end{bmatrix}$

This matrix is a reflection, you can prove it yourself if you want to.

This is outer product $[2,1]^T[2,1]$. So such a matrix has a very determined singular value decomposition. Only one singular value away from zero. You can then quite easily see it is projection on 5 times $[2,1]^T$. So both scaling and projection.


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