Tutorials › Linear Algebra › Powers Of Rotation-Scaling Blocks

Complex Eigenvalues and Complex Scalars · Tutorial 561 of 1000

Powers Of Rotation-Scaling Blocks

Use the scale and angle laws to compute integer powers of a rotation-scaling matrix, including when negative powers exist.

Intermediate 9 min read

What You'll Learn

  • Write a real rotation-scaling block using its scale factor and angle.
  • Multiply two blocks by combining their scales and adding their angles.
  • Compute positive, zero, and negative integer powers.
  • Identify when a negative power is defined and find the inverse block.
  • Check a computed power by multiplying matrices directly.
  • Avoid the common error of failing to multiply the angle by the exponent.

Scaling and Rotation in One Matrix

A rotation-scaling block describes two transformations of the plane at once: it scales every vector by a fixed factor and rotates it through a fixed angle. Repeating the transformation may look like a matrix-multiplication problem, but the block’s structure lets us track its scale and angle separately. This makes powers easier to compute and gives a direct way to check the result.

Throughout, the matrices act on column vectors in \(\mathbb{R}^2\). For a real angle \(\theta\), define the rotation matrix \( \mathbf{R}(\theta) \) and, for a scale factor \(r\geq 0\), the rotation-scaling block \( \mathbf{B}(r,\theta) \) by

$$ \mathbf{R}(\theta)= \begin{bmatrix} \cos\theta&-\sin\theta\\ \sin\theta&\cos\theta \end{bmatrix}, \qquad \mathbf{B}(r,\theta)=r\mathbf{R}(\theta)= \begin{bmatrix} r\cos\theta&-r\sin\theta\\ r\sin\theta&r\cos\theta \end{bmatrix}. $$

When \(r=0\), the block is the zero matrix regardless of the angle, so the angle does not describe a distinct transformation. When \(r>0\), the scale factor is positive and the angle determines the rotation; angles that differ by an integer multiple of \(2\pi\) give the same matrix. In the power formulas below, \(r^n\) is an ordinary scalar power.

Key rule: Multiplying rotation-scaling blocks multiplies their scale factors and adds their angles: \(\mathbf{B}(r,\theta)\mathbf{B}(s,\phi)=\mathbf{B}(rs,\theta+\phi)\) for \(r,s\geq 0\). In particular, for \(r>0\) and any integer \(n\), \(\mathbf{B}(r,\theta)^n=\mathbf{B}(r^n,n\theta)\). A negative power requires \(r>0\).

Why the Scale and Angle Combine This Way

The multiplication rule follows from multiplying the entries. For any angles \(\theta,\phi\), the trigonometric addition identities are \(\cos(\theta+\phi)=\cos\theta\cos\phi-\sin\theta\sin\phi\) and \(\sin(\theta+\phi)=\sin\theta\cos\phi+\cos\theta\sin\phi\). Multiplying the two rotation matrices gives exactly these combinations in its first column, with the corresponding negatives and rearranged terms in its second column.

Proof. Let \(c_\theta=\cos\theta\), \(s_\theta=\sin\theta\), and define \(c_\phi,s_\phi\) in the same way. Direct multiplication gives

$$ \mathbf{R}(\theta)\mathbf{R}(\phi) = \begin{bmatrix} c_\theta c_\phi-s_\theta s_\phi& -c_\theta s_\phi-s_\theta c_\phi\\ s_\theta c_\phi+c_\theta s_\phi& -s_\theta s_\phi+c_\theta c_\phi \end{bmatrix} =\mathbf{R}(\theta+\phi), $$

where the last equality uses the sine and cosine addition identities in each entry. Since scalar multiplication commutes with matrix multiplication, \(\mathbf{B}(r,\theta)\mathbf{B}(s,\phi)=rs\mathbf{R}(\theta)\mathbf{R}(\phi)=\mathbf{B}(rs,\theta+\phi)\). This also holds if either scale is zero, because both sides are then the zero matrix. \(\square\)

To obtain the power formula, apply the multiplication rule repeatedly. For exponent zero, the matrix power is defined by \(\mathbf{A}^0=\mathbf{I}_2\), including when \(\mathbf{A}\) is the zero matrix. For a negative exponent, \(\mathbf{A}^{-k}\) means \((\mathbf{A}^{-1})^k\), so it is defined only when \(\mathbf{A}\) has an inverse.

Theorem (Integer powers): If \(r>0\), then for every integer \(n\), \(\mathbf{B}(r,\theta)^n=\mathbf{B}(r^n,n\theta)\). If \(r=0\), then \(\mathbf{B}(0,\theta)^n=\mathbf{O}\) for positive integers \(n\), and \(\mathbf{B}(0,\theta)^0=\mathbf{I}_2\); negative powers are not defined.

Proof. First suppose \(r>0\). The formula for \(n=0\) holds because \(\mathbf{B}(r,\theta)^0=\mathbf{I}_2=\mathbf{B}(1,0)\). If it holds for some nonnegative integer \(k\), then the multiplication rule gives \(\mathbf{B}(r,\theta)^{k+1}=\mathbf{B}(r^k,k\theta)\mathbf{B}(r,\theta)=\mathbf{B}(r^{k+1},(k+1)\theta)\). By induction, the formula holds for every nonnegative integer \(n\).

For \(r>0\), the multiplication rule also gives \(\mathbf{B}(r,\theta)\mathbf{B}(r^{-1},-\theta)=\mathbf{B}(1,0)=\mathbf{I}_2\), and reversing the factors gives the same identity. Thus \(\mathbf{B}(r,\theta)^{-1}=\mathbf{B}(r^{-1},-\theta)\). For a positive integer \(k\), its \(k\)th power is \(\mathbf{B}(r^{-k},-k\theta)\) by the already proved nonnegative case applied to the inverse. This is the required formula for \(n=-k\). If \(r=0\), multiplying the zero matrix by itself any positive number of times gives the zero matrix, while exponent zero is defined to be \(\mathbf{I}_2\). The zero matrix has no inverse, so its negative powers are undefined. \(\square\)

Computing and Checking Powers

Worked Example: A Third Power with a Nontrivial Angle

Compute \(\mathbf{B}(2,\pi/3)^3\), then check the result by multiplying the original matrix three times.

Since \(\cos(\pi/3)=1/2\) and \(\sin(\pi/3)=\sqrt{3}/2\), the original block is

$$ \mathbf{A}=\mathbf{B}(2,\pi/3)= \begin{bmatrix} 1&-\sqrt{3}\\ \sqrt{3}&1 \end{bmatrix}. $$

The scale-angle rule predicts \(\mathbf{A}^3=\mathbf{B}(8,\pi)=\begin{bmatrix}-8&0\\0&-8\end{bmatrix}\). To verify by multiplication, first square the original matrix:

$$ \mathbf{A}^2= \begin{bmatrix} 1&-\sqrt{3}\\ \sqrt{3}&1 \end{bmatrix} \begin{bmatrix} 1&-\sqrt{3}\\ \sqrt{3}&1 \end{bmatrix} = \begin{bmatrix} 1-3&-\sqrt{3}-\sqrt{3}\\ \sqrt{3}+\sqrt{3}&-3+1 \end{bmatrix} = \begin{bmatrix} -2&-2\sqrt{3}\\ 2\sqrt{3}&-2 \end{bmatrix}. $$

Multiplying this result by \(\mathbf{A}\) gives

$$ \mathbf{A}^3= \begin{bmatrix} -2&-2\sqrt{3}\\ 2\sqrt{3}&-2 \end{bmatrix} \begin{bmatrix} 1&-\sqrt{3}\\ \sqrt{3}&1 \end{bmatrix} = \begin{bmatrix} -2-6&2\sqrt{3}-2\sqrt{3}\\ 2\sqrt{3}-2\sqrt{3}&-6-2 \end{bmatrix} = \begin{bmatrix} -8&0\\ 0&-8 \end{bmatrix}. $$

The scale is \(2^3=8\), and the angle is \(3(\pi/3)=\pi\), which gives the same matrix. The direct multiplication confirms every entry.

Worked Example: A Negative Power

Compute \(\mathbf{A}^{-2}\) for \(\mathbf{A}=\mathbf{B}(1/2,\pi/2)\), and check that the result multiplied by \(\mathbf{A}^2\) gives the identity.

The block is \(\mathbf{A}=\begin{bmatrix}0&-1/2\\1/2&0\end{bmatrix}\). Since its scale \(1/2\) is nonzero, the inverse exists. The inverse rule gives

$$ \mathbf{A}^{-1}=\mathbf{B}(2,-\pi/2)= \begin{bmatrix} 0&2\\ -2&0 \end{bmatrix}. $$

Squaring the inverse by entrywise multiplication yields

$$ \mathbf{A}^{-2}= \begin{bmatrix} 0&2\\ -2&0 \end{bmatrix} \begin{bmatrix} 0&2\\ -2&0 \end{bmatrix} = \begin{bmatrix} -4&0\\ 0&-4 \end{bmatrix}. $$

Check by squaring \(\mathbf{A}\) and multiplying back:

$$ \mathbf{A}^2= \begin{bmatrix} 0&-1/2\\ 1/2&0 \end{bmatrix} \begin{bmatrix} 0&-1/2\\ 1/2&0 \end{bmatrix} = \begin{bmatrix} -1/4&0\\ 0&-1/4 \end{bmatrix}, \qquad \mathbf{A}^2\mathbf{A}^{-2} = \begin{bmatrix} 1&0\\ 0&1 \end{bmatrix}. $$

The power rule gives the same answer: the scale is \((1/2)^{-2}=4\), and the angle is \(-2(\pi/2)=-\pi\), so \(\mathbf{B}(4,-\pi)=-4\mathbf{I}_2\).

Worked Example: A Negative Angle and Exponent Zero

Compute \(\mathbf{B}(3/2,-\pi/2)^3\), and also state the zeroth power of this block.

Since \(\cos(-\pi/2)=0\) and \(\sin(-\pi/2)=-1\), the original matrix is \(\mathbf{A}=\begin{bmatrix}0&3/2\\-3/2&0\end{bmatrix}\). The rule predicts scale \((3/2)^3=27/8\) and angle \(-3\pi/2\), which is equivalent to \(\pi/2\). Thus the predicted matrix is \(\begin{bmatrix}0&-27/8\\27/8&0\end{bmatrix}\).

Direct multiplication first gives

$$ \mathbf{A}^2= \begin{bmatrix} 0&3/2\\ -3/2&0 \end{bmatrix} \begin{bmatrix} 0&3/2\\ -3/2&0 \end{bmatrix} = \begin{bmatrix} -9/4&0\\ 0&-9/4 \end{bmatrix}. $$

Then

$$ \mathbf{A}^3= \begin{bmatrix} -9/4&0\\ 0&-9/4 \end{bmatrix} \begin{bmatrix} 0&3/2\\ -3/2&0 \end{bmatrix} = \begin{bmatrix} 0&-27/8\\ 27/8&0 \end{bmatrix}. $$

This agrees with the predicted block. The zeroth power is \(\mathbf{A}^0=\mathbf{I}_2=\begin{bmatrix}1&0\\0&1\end{bmatrix}\) by definition. As a direct check on the cube, multiplying \(\mathbf{A}^2\mathbf{A}\) above produces its entries \(0,-27/8,27/8,0\), respectively.

Common Mistakes and Practical Checks

Do not leave the angle unchanged when taking a power. For example, a tempting but incorrect calculation is \(\mathbf{B}(2,\pi/3)^2=\mathbf{B}(4,\pi/3)=\begin{bmatrix}2&-2\sqrt{3}\\2\sqrt{3}&2\end{bmatrix}\). It gets the scale \(4\) right but fails to double the angle. The corrected calculation is

$$ \mathbf{B}(2,\pi/3)^2 =\mathbf{B}(4,2\pi/3) = \begin{bmatrix} -2&-2\sqrt{3}\\ 2\sqrt{3}&-2 \end{bmatrix}. $$

Multiplying the original block by itself, as in the first worked example, verifies the corrected answer. The same angle rule applies to negative powers: the angle changes sign when the inverse is taken, because \(\mathbf{B}(r,\theta)^{-1}=\mathbf{B}(r^{-1},-\theta)\). Negative powers are not available when \(r=0\); applying a negative scalar exponent to the zero scale would not define a matrix inverse.

For a quick check, compare both parts of the result: the scale should be raised to the exponent, and the angle should be multiplied by it. Then multiply the proposed power by the original matrix when checking a positive power, or check that a proposed inverse multiplied by the original gives \(\mathbf{I}_2\). Angles may be simplified modulo \(2\pi\), but their multiples should be computed before that simplification.

These rules are useful whenever the same planar transformation is applied repeatedly. Instead of carrying out many matrix products, one can track how much the total scale changes and how much the total rotation accumulates. The multiplication proof explains why this shortcut works, while direct multiplication remains a reliable check on the entries and signs.

Check Your Understanding

Use the scale-angle rule, and check matrix answers by multiplication where requested.

  1. Compute \(\mathbf{B}(3,\pi/2)^2\) in block form and as a \(2\times2\) matrix.
  2. Find \(\mathbf{B}(2,-\pi/3)^{-1}\), and verify that multiplying it by \(\mathbf{B}(2,-\pi/3)\) gives \(\mathbf{I}_2\).
  3. What is \(\mathbf{B}(0,\pi/4)^3\)? Is \(\mathbf{B}(0,\pi/4)^{-1}\) defined?
  4. Correct the claim \(\mathbf{B}(1,\pi/6)^4=\mathbf{B}(1,\pi/6)\).
  5. Explain why the angle in \(\mathbf{B}(r,\theta)^n\) is \(n\theta\), not \(\theta\), when \(n\) is a positive integer.

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