TITLE: Describing a linear map geometrically
QUESTION [8 upvotes]: I have the following linear map $\mathbb{R}^2\to\mathbb{R}^2:$
$$\begin{pmatrix}x\\y\end{pmatrix}\mapsto \begin{pmatrix}9y-5x\\7y-4x\end{pmatrix}$$
I am asked to describe this geometrically. 
I can see what is happening: the upper left quadrant is being squeezed into the upper right quadrant, in between the lines $y=\frac{7}{9}x$ and $y=\frac{4}{5}x$. Likewise the lower right quadrant is scrambled in between the same lines in the lower left quadrant. The other quadrants are stretched in the obvious way.
I can't find the proper terms to describe this well though. A rotation followed by a squeeze? Thanks to anyone who responds.
(I can upload an image if anyone is interested, but the lines are so close that it will not be very clear)

REPLY [3 votes]: Suppose you have the map
$$
T_2
\begin{pmatrix}
x \\
y
\end{pmatrix}
=
\begin{pmatrix}
9y + x \\
y
\end{pmatrix}.
$$
Then this is called a horizontal shear. There is more information here: Shear mapping. Vertical shear is similarly defined. Now consider three additional maps
$$
T_1
\begin{pmatrix}
x \\
y
\end{pmatrix}
=
\begin{pmatrix}
-5x \\
y
\end{pmatrix} \\
T_3
\begin{pmatrix}
x \\
y
\end{pmatrix}
=
\begin{pmatrix}
x \\
-\frac{1}{5}y
\end{pmatrix} \\
T_4
\begin{pmatrix}
x \\
y
\end{pmatrix}
=
\begin{pmatrix}
x \\
y + \frac{4}{5}x
\end{pmatrix}.
$$
Then the given map is just
$$
T_4T_3T_2T_1
\begin{pmatrix}
x \\
y
\end{pmatrix}
=
\begin{pmatrix}
9y - 5x \\
7y - 4x
\end{pmatrix}.
$$
So the map is a combination of horizontal scaling followed by horizontal shear followed by vertical scaling and finally a vertical shear. Note that you can write transformations like this in terms of simpler transformations in many different ways. So the one I gave is just one possible way.

REPLY [2 votes]: A QR decomposition for $A=\begin{bmatrix}-5&9\\-4&7\end{bmatrix}$ is 
$$\begin{bmatrix}\frac{-5}{\sqrt{41}}&\frac{-4}{\sqrt{41}}\\\frac{4}{\sqrt{41}}&\frac{-5}{\sqrt{41}}\end{bmatrix}\begin{bmatrix}\sqrt{41}&\frac{-73}{\sqrt{41}}&\\0&\frac{1}{\sqrt{41}}&\end{bmatrix}$$
and we can additionally split out a dilation like so:
$$\begin{bmatrix}\frac{-5}{\sqrt{41}}&\frac{-4}{\sqrt{41}}\\\frac{4}{\sqrt{41}}&\frac{-5}{\sqrt{41}}\end{bmatrix}\begin{bmatrix}\sqrt{41}&0\\0&\sqrt{41}\end{bmatrix}\begin{bmatrix}1&\frac{-73}{41}&\\0&\frac{1}{41}\end{bmatrix}$$
In the last decomposition, let's call the matrices $R,D,T$ respectively.
This decomposition shows that $A$ is the triangular mapping $T$ followed by the dilation $D$ and then the rotation $R$.
The triangular mapping keeps the $x$ axis fixed and swings the $y$-axis around to a new direction. The dilation and reflection are self-explanatory, I hope. 
You can also note that the rotation is a proper rotation with no reflection, since its determinant is positive. The triangular mapping swings the $y$ axis out into the third quadrant. The dilation in this case expands the plane by a factor of about 6.7. The rotation is by about 141 degrees counterclockwise.

Analyzing it with singular value decomposition instead, we find two singular values of about $13.08$ and $0.08$. This says that although one direction is contracting, the orthogonal direction is expanding.