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Orthogonal Matrices and Isometries · Tutorial 710 of 1000

Products, Inverses, and Powers of Orthogonal Matrices

Learn why orthogonality is preserved by products and inverses, and use that structure to calculate matrix powers efficiently.

Intermediate 9 min read

What You'll Learn

  • Verify orthogonality from the transpose identity.
  • Prove that products and inverses of orthogonal matrices are orthogonal.
  • Apply induction to show that every integer power of an orthogonal matrix is orthogonal.
  • Compute powers of planar rotation matrices using angle addition.
  • Compute powers of reflection matrices and distinguish even from odd powers.
  • Identify and correct common errors in matrix-power calculations.

Orthogonality Survives Products and Inverses

Orthogonal matrices are useful not only because they preserve lengths and inner products, but also because their structure is retained under basic operations. If two orthogonal matrices can be composed, the product is orthogonal; reversing an orthogonal transformation also gives an orthogonal matrix. These facts let us calculate repeated transformations by studying powers of a single matrix, rather than checking each power from scratch.

Throughout this tutorial, matrices are real and square. A matrix \(\mathbf{Q}\in\mathbb{R}^{n\times n}\) is orthogonal when \(\mathbf{Q}^{\top}\mathbf{Q}=\mathbf{I}_n\). This condition says that the columns of \(\mathbf{Q}\) are orthonormal. For a square matrix, it also gives \(\mathbf{Q}\mathbf{Q}^{\top}=\mathbf{I}_n\) and \(\mathbf{Q}^{-1}=\mathbf{Q}^{\top}\): by the Invertible Matrix Theorem, a square matrix with a left inverse is invertible, and its left inverse must be its inverse.

Theorem (Closure under products and inverses): If \(\mathbf{A},\mathbf{B}\in\mathbb{R}^{n\times n}\) are orthogonal, then \(\mathbf{A}\mathbf{B}\) is orthogonal. Also, \(\mathbf{A}^{-1}\) is orthogonal.

Proof. For the product, use the transpose rule \((\mathbf{A}\mathbf{B})^{\top}=\mathbf{B}^{\top}\mathbf{A}^{\top}\), associativity of multiplication, and the orthogonality identities:

$$ (\mathbf{A}\mathbf{B})^{\top}(\mathbf{A}\mathbf{B}) =\mathbf{B}^{\top}\mathbf{A}^{\top}\mathbf{A}\mathbf{B} =\mathbf{B}^{\top}\mathbf{I}_n\mathbf{B} =\mathbf{B}^{\top}\mathbf{B} =\mathbf{I}_n. $$

Thus \(\mathbf{A}\mathbf{B}\) is orthogonal by definition. For the inverse, \(\mathbf{A}^{-1}=\mathbf{A}^{\top}\), so

$$ (\mathbf{A}^{-1})^{\top}\mathbf{A}^{-1} =(\mathbf{A}^{\top})^{\top}\mathbf{A}^{\top} =\mathbf{A}\mathbf{A}^{\top} =\mathbf{I}_n. $$

Therefore \(\mathbf{A}^{-1}\) is orthogonal as well. \(\square\)

The order of factors matters in matrix multiplication, but the proof does not require the matrices to commute: the middle factors \(\mathbf{A}^{\top}\mathbf{A}\) are adjacent and equal to \(\mathbf{I}_n\). The same attention to order is needed for inverses. In particular, \((\mathbf{A}\mathbf{B})^{-1}=\mathbf{B}^{-1}\mathbf{A}^{-1}\), not \(\mathbf{A}^{-1}\mathbf{B}^{-1}\). To check the order, multiply \(\mathbf{A}\mathbf{B}\) by \(\mathbf{B}^{-1}\mathbf{A}^{-1}\) on the right: the result is \(\mathbf{A}\mathbf{I}_n\mathbf{A}^{-1}=\mathbf{I}_n\).

Worked Example: A Product and Its Inverse

Let \(\mathbf{A}=\begin{bmatrix}0&-1\\1&0\end{bmatrix}\) and \(\mathbf{B}=\begin{bmatrix}1&0\\0&-1\end{bmatrix}\). Each matrix is orthogonal: direct multiplication gives \(\mathbf{A}^{\top}\mathbf{A}=\mathbf{I}_2\) and \(\mathbf{B}^{\top}\mathbf{B}=\mathbf{I}_2\). Compute their product and its inverse.

First multiply in the stated order:

$$ \mathbf{A}\mathbf{B} = \begin{bmatrix}0&-1\\1&0\end{bmatrix} \begin{bmatrix}1&0\\0&-1\end{bmatrix} = \begin{bmatrix}0&1\\1&0\end{bmatrix}. $$

Since this product is symmetric and its square is the identity, its inverse is itself. The transpose test confirms orthogonality:

$$ (\mathbf{A}\mathbf{B})^{\top}(\mathbf{A}\mathbf{B}) = \begin{bmatrix}0&1\\1&0\end{bmatrix} \begin{bmatrix}0&1\\1&0\end{bmatrix} = \begin{bmatrix}1&0\\0&1\end{bmatrix}. $$

For comparison, the inverse-order rule gives \((\mathbf{A}\mathbf{B})^{-1}=\mathbf{B}^{-1}\mathbf{A}^{-1}\). Here \(\mathbf{B}^{-1}=\mathbf{B}\) and \(\mathbf{A}^{-1}=\mathbf{A}^{\top}=\begin{bmatrix}0&1\\-1&0\end{bmatrix}\), so

$$ \mathbf{B}^{-1}\mathbf{A}^{-1} = \begin{bmatrix}1&0\\0&-1\end{bmatrix} \begin{bmatrix}0&1\\-1&0\end{bmatrix} = \begin{bmatrix}0&1\\1&0\end{bmatrix}. $$

Check: multiplying \(\mathbf{A}\mathbf{B}\) by the displayed inverse gives \(\begin{bmatrix}0&1\\1&0\end{bmatrix}^2=\mathbf{I}_2\), as required.

Integer Powers of an Orthogonal Matrix

For a square matrix \(\mathbf{Q}\), define \(\mathbf{Q}^0=\mathbf{I}_n\), and for a positive integer \(k\), define \(\mathbf{Q}^k\) as the product of \(k\) copies of \(\mathbf{Q}\). For a negative integer \(-k\), where \(k>0\) and \(\mathbf{Q}\) is invertible, define \(\mathbf{Q}^{-k}=(\mathbf{Q}^{-1})^k\). Every orthogonal matrix is invertible, so all of its integer powers are defined.

Theorem (Powers of an orthogonal matrix): If \(\mathbf{Q}\in\mathbb{R}^{n\times n}\) is orthogonal, then \(\mathbf{Q}^k\) is orthogonal for every integer \(k\).

Proof. First consider nonnegative integers. The identity \(\mathbf{Q}^0=\mathbf{I}_n\) is orthogonal, because \(\mathbf{I}_n^{\top}\mathbf{I}_n=\mathbf{I}_n\). Suppose \(\mathbf{Q}^k\) is orthogonal for some \(k\geq0\). By the product theorem, \(\mathbf{Q}^k\mathbf{Q}\) is orthogonal. Associativity and the definition of powers give \(\mathbf{Q}^k\mathbf{Q}=\mathbf{Q}^{k+1}\). Induction proves orthogonality for every \(k\geq0\). For \(k<0\), write \(k=-j\) with \(j>0\). The inverse theorem shows that \(\mathbf{Q}^{-1}\) is orthogonal, so the nonnegative case applied to \(\mathbf{Q}^{-1}\) shows that \((\mathbf{Q}^{-1})^j=\mathbf{Q}^{-j}\) is orthogonal. This covers every integer. \(\square\)

The proof explains why a power need not be checked by multiplying its transpose by itself each time. Once the original matrix is known to be orthogonal, closure under products and inverses does the work. This is a structural conclusion; a particular power can still be computed explicitly when the entries or the geometry are of interest.

Powers of Rotation Matrices

For a real angle \(\theta\), write the standard planar rotation matrix as

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

The columns are orthonormal because \(\cos^2\theta+\sin^2\theta=1\) and their dot product is \(-\cos\theta\sin\theta+\sin\theta\cos\theta=0\). Thus \(\mathbf{R}(\theta)\) is orthogonal. Multiplication combines angles: the entries of the product are given by the angle-addition identities, so \(\mathbf{R}(\alpha)\mathbf{R}(\beta)=\mathbf{R}(\alpha+\beta)\). In particular, repeated multiplication yields \(\mathbf{R}(\theta)^k=\mathbf{R}(k\theta)\) for nonnegative integers \(k\). For negative integers the same formula follows because \(\mathbf{R}(\theta)^{-1}=\mathbf{R}(-\theta)\).

Worked Example: A Rotation Through Sixty Degrees

Let \(\mathbf{R}=\mathbf{R}(\pi/3)=\begin{bmatrix}\tfrac12&-\tfrac{\sqrt3}{2}\\\tfrac{\sqrt3}{2}&\tfrac12\end{bmatrix}\). Compute \(\mathbf{R}^2\), \(\mathbf{R}^3\), and \(\mathbf{R}^{-1}\). The columns of \(\mathbf{R}\) have squared lengths \(\tfrac14+\tfrac34=1\), and their dot product is \(-\tfrac{\sqrt3}{4}+\tfrac{\sqrt3}{4}=0\), so the starting matrix is orthogonal.

Multiply \(\mathbf{R}\) by itself, displaying each entry:

$$ \mathbf{R}^2= \begin{bmatrix} \frac14-\frac34&-\frac{\sqrt3}{4}-\frac{\sqrt3}{4}\\ \frac{\sqrt3}{4}+\frac{\sqrt3}{4}&-\frac34+\frac14 \end{bmatrix} = \begin{bmatrix} -\frac12&-\frac{\sqrt3}{2}\\ \frac{\sqrt3}{2}&-\frac12 \end{bmatrix}. $$

Multiply once more. The top-left entry is \(-\tfrac14-\tfrac34=-1\), the top-right is \(\tfrac{\sqrt3}{4}-\tfrac{\sqrt3}{4}=0\), the bottom-left is \(-\tfrac{\sqrt3}{4}+\tfrac{\sqrt3}{4}=0\), and the bottom-right is \(-\tfrac34-\tfrac14=-1\). Hence \(\mathbf{R}^3=-\mathbf{I}_2\).

The inverse is the transpose:

$$ \mathbf{R}^{-1}=\mathbf{R}^{\top} =\begin{bmatrix}\frac12&\frac{\sqrt3}{2}\\-\frac{\sqrt3}{2}&\frac12\end{bmatrix}. $$

Check: multiplying \(\mathbf{R}\) by this inverse gives diagonal entries \(\tfrac14+\tfrac34=1\) and off-diagonal entries \(\tfrac{\sqrt3}{4}-\tfrac{\sqrt3}{4}=0\), so \(\mathbf{R}\mathbf{R}^{-1}=\mathbf{I}_2\). The computed powers also agree with the angle rule: \(2\pi/3\) gives the displayed \(\mathbf{R}^2\), and \(\pi\) gives \(-\mathbf{I}_2\).

Powers of Reflection Matrices

A reflection across a line through the origin making angle \(\theta\) with the positive horizontal axis has matrix

$$ \mathbf{S}(\theta)= \begin{bmatrix} \cos(2\theta)&\sin(2\theta)\\ \sin(2\theta)&-\cos(2\theta) \end{bmatrix}. $$

This matrix is symmetric. Its square is the identity: the diagonal entries of the square are \(\cos^2(2\theta)+\sin^2(2\theta)=1\), and both off-diagonal entries are \(\cos(2\theta)\sin(2\theta)-\sin(2\theta)\cos(2\theta)=0\). Therefore \(\mathbf{S}(\theta)^2=\mathbf{I}_2\), which also proves it is orthogonal, since \(\mathbf{S}(\theta)^{\top}\mathbf{S}(\theta)=\mathbf{I}_2\). It follows that even powers equal \(\mathbf{I}_2\), while odd powers equal \(\mathbf{S}(\theta)\): each successive pair of factors can be replaced by \(\mathbf{I}_2\). The same parity rule holds for negative integer powers because the inverse of this matrix is itself.

Worked Example: Repeated Reflection Across a Line

Take \(\theta=\pi/6\), so the line makes an angle of thirty degrees with the horizontal axis. The matrix is

$$ \mathbf{S}= \begin{bmatrix} \cos(\pi/3)&\sin(\pi/3)\\ \sin(\pi/3)&-\cos(\pi/3) \end{bmatrix} = \begin{bmatrix} \frac12&\frac{\sqrt3}{2}\\ \frac{\sqrt3}{2}&-\frac12 \end{bmatrix}. $$

Squaring gives top-left entry \(\tfrac14+\tfrac34=1\), top-right entry \(\tfrac{\sqrt3}{4}-\tfrac{\sqrt3}{4}=0\), bottom-left entry \(\tfrac{\sqrt3}{4}-\tfrac{\sqrt3}{4}=0\), and bottom-right entry \(\tfrac34+\tfrac14=1\). Therefore \(\mathbf{S}^2=\mathbf{I}_2\), and \(\mathbf{S}^3=\mathbf{S}^2\mathbf{S}=\mathbf{S}\). Also \(\mathbf{S}^{-1}=\mathbf{S}\), so \(\mathbf{S}^{-3}=\mathbf{S}^3=\mathbf{S}\).

Check: \(\mathbf{S}^2=\mathbf{I}_2\) verifies that the inverse is \(\mathbf{S}\), since multiplying \(\mathbf{S}\) by itself gives the identity. It also confirms that an even power returns the identity matrix and an odd power returns the original reflection matrix.

Common Pitfall: Reversing Factors or Assuming Powers Simplify

A frequent error is to reverse factors incorrectly when taking an inverse. For the matrices in the first example, \(\mathbf{A}^{-1}=\begin{bmatrix}0&1\\-1&0\end{bmatrix}\) and \(\mathbf{B}^{-1}=\mathbf{B}\). The incorrect order gives

$$ \mathbf{A}^{-1}\mathbf{B}^{-1} = \begin{bmatrix}0&1\\-1&0\end{bmatrix} \begin{bmatrix}1&0\\0&-1\end{bmatrix} = \begin{bmatrix}0&-1\\-1&0\end{bmatrix}, $$

which is not \((\mathbf{A}\mathbf{B})^{-1}=\begin{bmatrix}0&1\\1&0\end{bmatrix}\). The correction is to reverse both the factors and their order: \((\mathbf{A}\mathbf{B})^{-1}=\mathbf{B}^{-1}\mathbf{A}^{-1}\). Multiplying the corrected expression gives the actual inverse and verifies the rule.

Another error is to treat every orthogonal matrix like a reflection and conclude that its square is the identity. The rotation in the previous example contradicts this: \(\mathbf{R}^2=\begin{bmatrix}-\tfrac12&-\tfrac{\sqrt3}{2}\\\tfrac{\sqrt3}{2}&-\tfrac12\end{bmatrix}\ne\mathbf{I}_2\). Orthogonality guarantees that powers remain orthogonal; it does not say those powers equal the identity. The special parity pattern for reflection powers follows from the additional fact \(\mathbf{S}^2=\mathbf{I}_2\).

These closure properties are valuable whenever a transformation is applied repeatedly or when several length-preserving transformations are composed. They provide a quick structural check: a product of orthogonal matrices must be orthogonal, and an inverse must be orthogonal. For actual entries, use the relevant structure as well—angle addition for the rotation matrices above, and the square-equals-identity relation for the displayed reflections.

Check Your Understanding

Use the transpose criterion and the power rules to justify each conclusion.

  1. If \(\mathbf{A}\) and \(\mathbf{B}\) are orthogonal matrices of the same size, expand \((\mathbf{A}\mathbf{B})^{\top}(\mathbf{A}\mathbf{B})\) and identify the step that uses orthogonality.
  2. For \(\mathbf{R}(\pi/2)=\begin{bmatrix}0&-1\\1&0\end{bmatrix}\), compute \(\mathbf{R}(\pi/2)^4\) using the angle rule.
  3. For an orthogonal matrix \(\mathbf{Q}\), explain why \(\mathbf{Q}^{-3}\) is orthogonal.
  4. For the reflection matrix \(\mathbf{S}\) in the worked example, compute \(\mathbf{S}^{2025}\) and \(\mathbf{S}^{-2025}\).
  5. State the correct order of factors in the inverse of a product and verify the order by multiplying the product by the proposed inverse.

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