# What two transformations are described by $A=\begin{bmatrix} 2 & 4 \\\ 1 & 2 \end{bmatrix}$

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.

• Any linear transformation is a composition of two linear transformations. You can take $B=A,C=I$. – Kavi Rama Murthy Apr 18 at 7:36
• @KaviRamaMurthy Yes, that is certainly true but my exercise restricts me from using $C=I$. I apologize for not clarifying. – 高田航 Apr 18 at 7:44
• Then take $B=2A$, $C=\frac 1 2 I$. – Kavi Rama Murthy Apr 18 at 7:45
• 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. – 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.