Reading the Matrix as a Transformation of the Plane
Consider the real matrix \(\mathbf{M}=\begin{bmatrix}a&-b\\b&a\end{bmatrix}\). Its entries have a useful geometric meaning: the matrix takes every vector in the plane, scales its length by the same nonnegative factor, and rotates it through an angle determined by \(a\) and \(b\). Connecting this action to the complex number \(a+ib\) makes both the scale and the angle visible.
Write a vector \(\mathbf{x}=\begin{bmatrix}x\\y\end{bmatrix}\in\mathbb{R}^2\) as the complex number \(z=x+iy\). This correspondence is just a way to represent the same two real coordinates. Multiplying by \(\mathbf{M}\) gives
The coordinates on the right are the real and imaginary parts of \((a+ib)(x+iy)\). Thus the matrix acts on the plane exactly as multiplication by \(a+ib\) acts on complex numbers. The geometric effect of this multiplication follows from polar form.
The Scale-and-Rotation Factorization
When \((a,b)\ne(0,0)\), the value \(r=\sqrt{a^2+b^2}\) is positive. Substituting \(a=r\cos\theta\) and \(b=r\sin\theta\) into the entries of \(\mathbf{M}\) gives a factorization into a positive scale and a rotation:
Proof. Suppose first that \((a,b)\ne(0,0)\). By the polar representation of \(a+ib\), there are \(r=\sqrt{a^2+b^2}>0\) and an argument \(\theta\) with \(a=r\cos\theta\) and \(b=r\sin\theta\). Substitution into the entries gives \(\mathbf{M}=r\mathbf{R}_{\theta}\). Therefore \(\mathbf{M}\mathbf{x}=r(\mathbf{R}_{\theta}\mathbf{x})\), which is a rotation followed by scaling. To verify the length claim directly, for \(\mathbf{x}=\begin{bmatrix}x\\y\end{bmatrix}\) we have \(\mathbf{M}\mathbf{x}=\begin{bmatrix}ax-by\\bx+ay\end{bmatrix}\), so \[ \|\mathbf{M}\mathbf{x}\|^2=(ax-by)^2+(bx+ay)^2 =(a^2+b^2)(x^2+y^2)=r^2\|\mathbf{x}\|^2. \] Both sides’ lengths are nonnegative and \(r>0\), hence \(\|\mathbf{M}\mathbf{x}\|=r\|\mathbf{x}\|\). If \(a=b=0\), every entry of \(\mathbf{M}\) is zero, so \(\mathbf{M}\mathbf{x}=\mathbf{0}\) for all \(\mathbf{x}\). In this case \(r=0\), while \(a+ib=0\) has no argument, so no angle is specified. \(\square\)
For a nonzero input vector, the theorem says not only that its length changes predictably, but also that its direction changes by the same angle \(\theta\), regardless of the vector’s initial direction. If the input is \(\mathbf{0}\), its image remains \(\mathbf{0}\); a direction or angle for that vector is not defined. The matrix itself has a rotation angle only when its scale is nonzero.
Examples: Reading the Angle and Scale
Worked Example: A Positive Quarter-Turn and Doubling
Take \(a=0\) and \(b=2\). Then \(\mathbf{M}=\begin{bmatrix}0&-2\\2&0\end{bmatrix}\). Find its scale and angle, and check its action on \(\mathbf{x}=\begin{bmatrix}1\\0\end{bmatrix}\).
The scale is \(r=\sqrt{0^2+2^2}=2\). The complex number is \(a+ib=2i\), whose principal argument is \(\theta=\frac{\pi}{2}\), since \(\cos(\frac{\pi}{2})=0\) and \(\sin(\frac{\pi}{2})=1\). Hence the matrix is twice a counterclockwise quarter-turn. Multiplication gives
Check: \(\mathbf{R}_{\pi/2}\begin{bmatrix}1\\0\end{bmatrix}=\begin{bmatrix}0\\1\end{bmatrix}\), and multiplying this result by \(2\) gives \(\begin{bmatrix}0\\2\end{bmatrix}\), as the matrix product does. Its length changes from \(1\) to \(2\).
Worked Example: A Scale of Five in the First Quadrant
Take \(a=3\) and \(b=4\), and apply \(\mathbf{M}\) to \(\mathbf{x}=\begin{bmatrix}1\\2\end{bmatrix}\). Determine the scale and angle, and verify the factorized action.
Here \(r=\sqrt{3^2+4^2}=5\). Since \(a/r=3/5\) and \(b/r=4/5\), an angle \(\theta\) has \(\cos\theta=3/5\) and \(\sin\theta=4/5\); it lies in the first quadrant. The matrix is \(\mathbf{M}=\begin{bmatrix}3&-4\\4&3\end{bmatrix}\). Direct multiplication gives
To check the rotation first, use the stated sine and cosine:
Scaling by \(5\) yields \(\begin{bmatrix}-5\\10\end{bmatrix}\), matching the direct matrix product. Also, the original length is \(\sqrt{1^2+2^2}=\sqrt{5}\), while the image length is \(\sqrt{(-5)^2+10^2}=5\sqrt{5}\), exactly five times as large.
Worked Example: A Negative Real Multiplier
Take \(a=-3\), \(b=0\), and \(\mathbf{x}=\begin{bmatrix}2\\-1\end{bmatrix}\). Find the scale and angle and verify the result. This case shows why a negative \(a\) does not mean a negative scale.
The scale is \(r=\sqrt{(-3)^2+0^2}=3\), and \(a+ib=-3\) has principal argument \(\theta=\pi\). Thus \(\mathbf{M}=\begin{bmatrix}-3&0\\0&-3\end{bmatrix}=3\mathbf{R}_{\pi}\). Direct multiplication gives
Check: \(\mathbf{R}_{\pi}\mathbf{x}=\begin{bmatrix}-1&0\\0&-1\end{bmatrix}\begin{bmatrix}2\\-1\end{bmatrix}=\begin{bmatrix}-2\\1\end{bmatrix}\). Multiplying by \(3\) gives \(\begin{bmatrix}-6\\3\end{bmatrix}\), the direct result. The length changes from \(\sqrt{5}\) to \(3\sqrt{5}\), so the scale is positive \(3\); the reversal of direction comes from the half-turn.
Signs, Quadrants, and the Zero Case
The signs of \(a\) and \(b\) locate \(a+ib\) in the complex plane and therefore determine which argument represents the rotation. The principal argument convention keeps the reported angle in \((-\pi,\pi]\). The signs do not replace checking the quadrant: for instance, \(a<0,b>0\) gives an angle in quadrant II, while \(a<0,b<0\) gives a principal angle in quadrant III, reported as a negative angle.
| Parameters | Scale and principal angle | Geometric reading |
|---|---|---|
| \(a>0,\ b=0\) | \(r=a,\ \theta=0\) | Positive scaling, with no turn |
| \(a<0,\ b=0\) | \(r=|a|,\ \theta=\pi\) | Positive scaling and a half-turn |
| \(a=0,\ b>0\) | \(r=b,\ \theta=\frac{\pi}{2}\) | Positive scaling and a counterclockwise quarter-turn |
| \(a=0,\ b<0\) | \(r=|b|,\ \theta=-\frac{\pi}{2}\) | Positive scaling and a clockwise quarter-turn |
| \(a=b=0\) | \(r=0\), angle undefined | Every vector maps to \(\mathbf{0}\) |
For other nonzero parameter pairs, use \(r=\sqrt{a^2+b^2}\) and choose the angle whose cosine is \(a/r\) and sine is \(b/r\), taking the signs into account. The expression \(\arctan(b/a)\) alone can give the wrong quadrant, and it is not defined when \(a=0\). The argument is the safer description because it records both coordinates’ signs.
The zero case needs separate attention. If \(a=b=0\), then the matrix is \(\mathbf{O}\) and there is no direction in the complex plane from \(0\) to identify as an angle. One may write \(\mathbf{O}=0\mathbf{R}_{\theta}\) for any \(\theta\), but that does not assign a unique rotation to the zero matrix: multiplying by zero erases every input direction. It is more precise to say that the matrix collapses the plane to the origin and has no determined rotation angle.
Common Sign Error
A frequent mistake is to read the negative entry in the top-right corner as evidence that positive \(b\) produces a clockwise turn. The sign must be interpreted together with the bottom-left entry and the action on a vector. For \(a=0,b=1\), the matrix is \(\begin{bmatrix}0&-1\\1&0\end{bmatrix}\). The mistaken calculation might claim that \(\begin{bmatrix}1\\0\end{bmatrix}\) maps to \(\begin{bmatrix}0\\-1\end{bmatrix}\), as if the turn were clockwise. But actual multiplication gives
The first standard basis vector points to the right, and its image points upward, so this is a counterclockwise quarter-turn. The correction is to use the full matrix pattern: its columns are the images of the standard basis vectors, and for \(b>0,a=0\), the first column is \(\begin{bmatrix}0\\b\end{bmatrix}\), pointing upward. The second column \(\begin{bmatrix}-b\\0\end{bmatrix}\) points left, consistently giving the same counterclockwise rotation.
Interpreting \(\mathbf{M}\) as \(r\mathbf{R}_{\theta}\) is useful because it separates two independent geometric effects: the nonnegative scale \(r\) changes lengths uniformly, while the argument \(\theta\) changes directions uniformly. The factorization applies to every nonzero pair \((a,b)\), including the axis cases, while the zero pair must be handled separately because its angle is undefined.
Check Your Understanding
Use the scale-and-rotation interpretation, and verify any claimed action by multiplying the matrix by the vector.
- For \(a=0,b=-3\), find the scale and principal angle. Where does \(\begin{bmatrix}1\\0\end{bmatrix}\) go?
- For \(a=-4,b=0\), explain why the scale is positive even though \(a\) is negative.
- For \(a=1,b=-1\), find \(r\) and the principal argument of \(a+ib\).
- What transformation does the matrix represent when \(a=b=0\), and why is its rotation angle not determined?
- Show directly from the factorization that \(\mathbf{M}\mathbf{x}\) has length \(r\|\mathbf{x}\|\) when \((a,b)\ne(0,0)\).
