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For a line through the origin making angle $\theta$ with the positive $x$-axis, its reflection matrix isIts determinant is $-1$. An orthogonal matrix has determinant $\pm 1$. If its determinant is $-1$, write its first column as $(\cos \phi ,\sin \phi {)}^T$; orthonormality and the determinant then force the second column to be $(\sin \phi ,-\cos \phi {)}^T$. The matrix is $S_{\phi /2}$, hence is one Euclidean reflection. If its determinant is $+1$, the second column is $(-\sin \phi ,\cos \phi {)}^T$, so it is a rotation matrix. Direct multiplication givesThe identity may be taken as an empty product or as two identical reflections. Thus at most two origin-line reflections suffice in every case.

For a nonzero vector $b=dn$ with $d=|b|$ and unit normal $n$, reflection in the affine line $n\cdot x=c$ is $s_c(x)=x+2(c-n\cdot x)n$. Consequently $s_{d/2}{s}_0(x)=x+dn=x+b$. A translation is therefore a product of two parallel-line reflections. Combining this with the given decomposition of a Euclidean isometry as translation followed by an orthogonal map proves that every plane isometry is a finite product of affine-line reflections. The affine lines need not pass through the origin. In fact three suffice: if an isometry $f$ sends zero to $b\ne 0$, reflection in the perpendicular bisector of $0,b$ takes $b$ to zero. Composing that reflection with $f$ gives an isometry fixing zero, to which the two-reflection linear result applies. This is the planar finite reflection decomposition of a Euclidean isometry.

For an example that genuinely needs three, take the glide reflectionIt has no fixed point, so it is not one reflection. Its orthogonal part has determinant $-1$, whereas the product of two reflections has determinant $+1$, so it is not two reflections; it is not the identity either. Explicitly it is reflection in $y=0$ followed by reflection in $x=0$ and then in $x=1/2$. Thus its minimum reflection count is exactly three.

The orthogonal reflection in $x=y$ exchanges the first two coordinates and leaves the third unchanged. Its fixed subspace is the plane, while its normal $(1,-1,0)$ changes sign. ThereforeThis reflection matrix has determinant $-1$, as expected.

All the transformations below fix the origin. A planar rotation preserves lengths and angles and turns every vector through one common oriented angle. Its rotation matrix isAn orthogonal reflection in a line fixes that line and reverses its perpendicular direction; reflection in the horizontal axis has reflection matrix $\mathrm{diag}(1,-1)$. A uniform dilation multiplies every vector by the same positive scale $s$, represented by $sI$; for example $2I$ doubles lengths. A shear mapping fixes one line pointwise and shifts vectors parallel to that line by an amount proportional to their transverse coordinate. A horizontal shear mapping has matrixwith $S_1$ a nontrivial example.

The first given matrix is the clockwise planar rotation through $\pi /4$, namely $A=R_{-\pi /4}$. Direct multiplication givesThus $C$ is a horizontal shear mapping of strength $2$. Because $A^{-1}=R_{\pi /4}$,So $B$ is the composition of that shear mapping followed by an anticlockwise planar rotation through $\pi /4$; order matters. It is not individually one of the four elementary types under these definitions. In particular, its determinant is $1$, but it is not an orthogonal matrix, excluding both a planar rotation and an orthogonal reflection. It is not a scalar matrix, excluding a uniform dilation; its eigenvalues $\sqrt{2}+1$ and $\sqrt{2}-1$ exclude a pure shear mapping, whose eigenvalues are both $1$. As a symmetric positive-definite matrix, $B$ can also be described as stretching two perpendicular principal directions by these reciprocal factors, preserving area.

Denote the four displayed matrices, from left to right, by $T_1,T_2,T_3,T_4$. An orthogonal matrix has an orthonormal set of columns, equivalently $T^{T}T=I$. Direct column dot products give $T_j^T{T}_j=I$ for $j=1,2,4$. The first column of $T_3$ has squared Euclidean norm $(1+6+1)/6=4/3$, so the third matrix is the unique nonorthogonal matrix.

For $T_1$, the determinant is $1$. A nonidentity orthogonal transformation of ${\mathbb{R}}^3$ with determinant $1$ is a rotation: it has a fixed axis and restricts to a plane rotation on its orthogonal complement. Consequently $T_1$ is the rotation.

For $T_2$, calculation givesA plane reflection fixes its entire plane, and therefore has eigenvalue $1$. Since $T_2-I$ is invertible, $T_2$ is not a plane reflection. Thus $T_2$ is the combination of a rotation and a reflection. More explicitly, $(1,-2,0{)}^T$ is an eigenvector with eigenvalue $-1$. On the perpendicular plane, $T_2$ is a rotation with $\cos \varphi =2/3$, because $\mathrm{tr}{T}_2=-1+2\cos \varphi =1/3$. This gives a three-dimensional improper orthogonal transformation.

For $T_4$, set $n=(1,2,2{)}^T/3$. ThenThis reflection matrix reverses the normal $n$ and fixes every vector perpendicular to $n$. Therefore $T_4$ is reflection in the plane $x+2y+2z=0$. In particular, $\det T_4=-1$ and $\det (T_4-I)=0$. Finally $T_3$ represents none of the three listed orthogonal transformations, since each preserves the Euclidean norm.

Choose the $z$-axis along the line of intersection of the two planes. Their unit normals lie in the $xy$-plane; after choosing the $x$-axis, takewhere $\theta$ is the oriented angle between the planes. The two reflection matrices restrict toon the $xy$-plane and both fix the $z$-axis. Their product isThus the composition of two plane reflections is a rotation about the intersection line through angle $2\theta$.

Writing $n$ as a column vector, the reflection isso its reflection matrix is

Among the two orthogonal matrices from part (d)(i), $d=-1$ gives $L=-I$, a rotation matrix through angle $\pi$. For $d=1$, the determinant is $-1$, so the matrix reverses orientation and is a reflection rather than a rotation:It is the reflection matrix across the $y$ axis.

A unit vector along the line is $v=(3,4{)}^T/5$. Reflection fixes its component parallel to $v$ and negates its perpendicular component, so the reflection matrix isIndeed $H(3,4{)}^T=(3,4{)}^T$, $H(-4,3{)}^T=-(-4,3{)}^T$ and $H^2=I$. These characterize the orthogonal reflection in that line.

A unit vector along the line is $d=(3,4{)}^{\top }/5$. Reflection preserves the orthogonal projection $d{d}^{\top }u$ onto that line and negates its perpendicular component. Therefore $Au=d{d}^{\top }u-(u-d{d}^{\top }u)$ andAs checks, $A(3,4{)}^{\top }=(3,4{)}^{\top }$, $A(-4,3{)}^{\top }=-(-4,3{)}^{\top }$, and $A^2=I$. Its determinant is minus one, as expected for a planar reflection matrix.

For a nonzero real normal vector $n$, orthogonal reflection in $n^{T}x=d$ has the displayed affine map. It is $x\mapsto Hx+t$ with $H=I-2n{n}^T/(n^{T}n)$ and $t=2dn/(n^{T}n)$. In homogeneous coordinates its matrix isEvery point on the hyperplane is fixed, and the signed normal displacement changes sign. Choosing any point $p$ on the hyperplane gives $t=p-Hp$, so translating by $-p$, applying the reflection matrix $H$ and translating back gives the same transformation.

A rotation matrix is an orthogonal matrix with determinant $1$.

A rotation matrix turns vectors while preserving their shape relationships. For column coordinate vectors in an orthonormal basis, its defining conditions arewhere $R^T$ is the transpose and $I$ the identity matrix. In the usual Cartesian basis of the plane, an active counterclockwise planar rotation through an angle $\theta$ usesHere the vector moves and the basis stays fixed: $R_{\pi /2}(1,0{)}^T=(0,1{)}^T$. Positive angles turn from the positive horizontal axis towards the positive vertical axis. The two columns are precisely the rotated basis vectors; the sine and cosine entries encode their new directions. Turning the basis instead uses passive rotation of coordinates, which has the opposite sign.

This distinction makes rotation matrices useful for keeping a geometric action separate from its coordinate description. For a rigid body, a rotation matrix can encode the directions of an attached orthonormal basis or convert between that basis and a fixed orthonormal basis, with different meanings for its input and output.^[ref]

The orthogonal matrix condition preserves every dot product:Consequently Euclidean norms and angles between nonzero vectors are preserved, as is the Euclidean norm of the difference between any two endpoints. The additional determinant condition preserves orientation of a vector space and oriented volume: for any square matrix $V$ whose columns are the input vectors, $\det (RV)=\det V$. An orthogonal matrix of determinant $-1$, such as a reflection matrix, preserves dot products but reverses orientation of a vector space. The matrix inverse is especially simple: $R^{-1}=R^T$.^[ref]