Orthogonal Matrix¶

Definition¶

A real $n\times n$ matrix $Q$ is orthogonal if

Q^{\top}Q=I,

where $I$ is the $n\times n$ identity matrix. Equivalently, $Q$ is invertible with $Q^{-1}=Q^{\top}$ $Q^{⊤}$ , so also $QQ^{\top}=I$ $I$ .

Interpretation¶

The $(i,j)$ entry of $Q^{\top}Q$ is the inner product $\boldsymbol{q}_i\cdot\boldsymbol{q}_j$ of columns $i$ and $j$ of $Q$ . So $Q^{\top}Q=I$ says exactly that the columns form an orthonormal basis of $\mathbb{R}^n$ , and $QQ^{\top}=I$ says the same of the rows. The name is traditional: the columns must be orthonormal, not merely orthogonal. The diagonal matrix $\operatorname{diag}(2,1)$ has orthogonal columns but is not an orthogonal matrix.

Geometrically, $\boldsymbol{x}\mapsto Q\boldsymbol{x}$ is a rigid motion that fixes the origin: it preserves lengths, angles and distances. In $\mathbb{R}^2$ every orthogonal matrix is a rotation or a reflection in a line through the origin.

Two panels: a rotation and a reflection each send the standard basis e1, e2 to orthonormal columns q1, q2 on the unit circle; the reflection's mirror line is drawn dashed.

Both panels start from $\boldsymbol{e}_1,\boldsymbol{e}_2$ (grey); the images $\boldsymbol{q}_1=Q\boldsymbol{e}_1$ (blue) and $\boldsymbol{q}_2=Q\boldsymbol{e}_2$ (red) are the columns of $Q$ , and the shaded square is the image of the unit square. On the right $Q$ is the matrix $P=\tfrac{1}{\sqrt{2}}\begin{pmatrix}1&1\\1&-1\end{pmatrix}$ from the Examples, which is $S_\theta$ with $\theta=45^\circ$ $4 5^{\circ}$ : it reflects in the line at $22.5^\circ$ (orange) and has $\det P=-1$ $- 1$ . On the left the sign of its second column is swapped, which gives the rotation $R_{45^\circ}=\tfrac{1}{\sqrt{2}}\begin{pmatrix}1&-1\\1&1\end{pmatrix}$ with determinant $1$ . In both cases the columns stay on the unit circle and at a right angle, so lengths and angles are preserved; only the orientation of the square differs, and that is what the sign of $\det Q$ records.

Turn $\theta$ and drag $\boldsymbol{x}$ : under $R_\theta$ or $S_\theta$ the unit circle maps onto itself and $\|Q\boldsymbol{x}\|=\|\boldsymbol{x}\|$ $∥ x ∥$ , while $\det Q=-1$ shows up as a mirrored F. Move the length $k$ of the second column away from $1$ : the columns stay orthogonal, but $Q^{\top}Q\neq I$ $I$ , the circle becomes an ellipse and lengths change. Switch on $Q^{\top}$ to see that it undoes $Q$ exactly when $Q$ is orthogonal.

Orthogonal matrices make changes of basis cheap: if the new basis is orthonormal, its transformation matrix $Q$ is orthogonal and the inverse needed for new coordinates is just the transpose, $\boldsymbol{y}=Q^{\top}\boldsymbol{x}$ $Q^{⊤} x$ . This is why they appear in central factorizations. A symmetric matrix can be diagonalized by an orthogonal matrix, $A=PDP^{\top}$ with orthonormal eigenvectors as the columns of $P$ . In Week 7, the singular value decomposition $A=U\Sigma V^{\top}$ of an arbitrary matrix also uses orthogonal $U$ and $V$ .

A rectangular $n\times k$ matrix $Q$ with $k<n$ and orthonormal columns satisfies $Q^{\top}Q=I_k$ but not $QQ^{\top}=I_n$ $I_{n}$ ; $QQ^{\top}$ is then the orthogonal projection onto its column space. The word "orthogonal matrix" is reserved for the square case.

Key Results¶

$Q$ preserves inner products and lengths: $(Q\boldsymbol{x})\cdot(Q\boldsymbol{y})=\boldsymbol{x}^{\top}Q^{\top}Q\boldsymbol{y}=\boldsymbol{x}\cdot\boldsymbol{y}$ $y$ , hence $\|Q\boldsymbol{x}\|=\|\boldsymbol{x}\|$ $∥ x ∥$ . Conversely, a square matrix with $\|Q\boldsymbol{x}\|=\|\boldsymbol{x}\|$ for all $\boldsymbol{x}$ is orthogonal.
$\det Q=\pm1$ $\pm 1$ , since $(\det Q)^2=\det(Q^{\top}Q)=\det I=1$ by the product rule for the determinant. In $\mathbb{R}^2$ and $\mathbb{R}^3$ , $\det Q=1$ gives a rotation and $\det Q=-1$ includes a reflection. The converse fails: $\begin{pmatrix}1&1\\0&1\end{pmatrix}$ has determinant $1$ but is not orthogonal, since its second column has length $\sqrt{2}$ .
Every (possibly complex) eigenvalue $\lambda$ of $Q$ has $|\lambda|=1$ $1$ ; the only possible real eigenvalues are $1$ and $-1$ .
Products, inverses and transposes of orthogonal matrices are orthogonal: $(Q_1Q_2)^{\top}Q_1Q_2=Q_2^{\top}Q_2=I$ $I$ . Sums generally are not: $I+I=2I$ is not orthogonal.
A linear system $Q\boldsymbol{x}=\boldsymbol{b}$ with $Q$ orthogonal is solved by $\boldsymbol{x}=Q^{\top}\boldsymbol{b}$ $Q^{⊤} b$ , without elimination; equivalently, $x_i=\boldsymbol{q}_i\cdot\boldsymbol{b}$ $b$ .
Every permutation matrix is orthogonal.
Applying the Gram-Schmidt algorithm to any basis of $\mathbb{R}^n$ and placing the resulting vectors as columns produces an orthogonal matrix.

Examples¶

Rotations and reflections in $\mathbb{R}^2$ . For an angle $\theta$ ,

\begin{aligned} R_\theta&=\begin{pmatrix}\cos\theta&-\sin\theta\\\sin\theta&\cos\theta\end{pmatrix},\\ S_\theta&=\begin{pmatrix}\cos\theta&\sin\theta\\\sin\theta&-\cos\theta\end{pmatrix}. \end{aligned}

Both have orthonormal columns because $\cos^2\theta+\sin^2\theta=1$ $1$ . $R_\theta$ rotates by $\theta$ and has $\det R_\theta=1$ $1$ ; $S_\theta$ reflects in the line at angle $\theta/2$ and has $\det S_\theta=-1$ $- 1$ .

For example, $R_{\pi/2}=\begin{pmatrix}0&-1\\1&0\end{pmatrix}$ maps $(3,4)^{\top}$ to $(-4,3)^{\top}$ ; both vectors have length $5$ , and their dot product $-12+12=0$ confirms the right angle.

Checking a $3\times3$ matrix. Let

Q=\frac{1}{3}\begin{pmatrix}1&2&2\\2&1&-2\\2&-2&1\end{pmatrix}.

Each column has squared length $\tfrac{1}{9}(1+4+4)=1$ $1$ , and the columns are orthogonal, for instance $\tfrac{1}{9}(1\cdot2+2\cdot1+2\cdot(-2))=0$ $0$ ; so $Q^{\top}Q=I$ and $Q^{-1}=Q^{\top}=Q$ $Q$ . Expanding the determinant gives $\det Q=\tfrac{1}{27}(-27)=-1$ $- 1$ . In fact $Q\boldsymbol{u}=-\boldsymbol{u}$ for $\boldsymbol{u}=(1,-1,-1)^{\top}$ and $Q\boldsymbol{w}=\boldsymbol{w}$ for every $\boldsymbol{w}\perp\boldsymbol{u}$ $u$ : $Q$ is the reflection in the plane orthogonal to $\boldsymbol{u}$ .

Diagonalizing a symmetric matrix. For $A=\begin{pmatrix}2&1\\1&2\end{pmatrix}$ $(2112)$ , the eigenvalues are $3$ and $1$ with unit eigenvectors $\tfrac{1}{\sqrt{2}}(1,1)^{\top}$ and $\tfrac{1}{\sqrt{2}}(1,-1)^{\top}$ . With

P=\frac{1}{\sqrt{2}}\begin{pmatrix}1&1\\1&-1\end{pmatrix},

$P$ is orthogonal and $A=PDP^{\top}$ with $D=\operatorname{diag}(3,1)$ $diag (3, 1)$ : no matrix inversion is needed. Here $\det P=-1$ $- 1$ ; swapping the sign of one column gives a rotation instead.

Prerequisites¶

Sources¶

Math V: Advanced Linear Algebra · Week 6 (placement provisional; not yet checked against the slides)

Aliases: orthogonal matrices, orthogonal transformation, orthonormal matrix Keywords: Q^T Q = I, Q^{-1} = Q^T, det Q = ±1, orthonormal columns, rotation matrix, reflection matrix, length-preserving, isometry Symbols: lambda, pi, real numbers, sigma, square root, theta

Last updated: 17 October 2026 21:53