Why the Plane Has Two Orthogonal Forms
An orthogonal matrix preserves lengths and angles, so in the plane its columns must be unit vectors at right angles to each other. Once the first column is chosen, there are exactly two possible unit vectors perpendicular to it: one on each side of the first column. Those two choices produce the two forms that classify every real \(2\times2\) orthogonal matrix. The determinant distinguishes the forms: one has determinant \(1\), and the other has determinant \(-1\).
For a real square matrix \(\mathbf{Q}\), orthogonality means \(\mathbf{Q}^{\top}\mathbf{Q}=\mathbf{I}\). The columns of \(\mathbf{Q}\) are therefore orthonormal: each has length \(1\), and distinct columns have dot product \(0\). In this tutorial, a rotation matrix means a planar orthogonal matrix with determinant \(1\), and a reflection matrix means a planar orthogonal matrix with determinant \(-1\). The classification below derives their forms; it does not require describing rotations by angles or finding reflection lines.
The equivalence between the matrix equation and the column condition follows by multiplying \(\mathbf{Q}^{\top}\mathbf{Q}\): its diagonal entries are the squared lengths of the columns, and its off-diagonal entries are their dot product. Also, taking determinants in \(\mathbf{Q}^{\top}\mathbf{Q}=\mathbf{I}_2\) gives \((\det\mathbf{Q})^2=1\), so \(\det\mathbf{Q}\) must be \(1\) or \(-1\).
Deriving the Two Forms
Write the columns of \(\mathbf{Q}\) as \(\mathbf{u}=(x,y)^{\top}\) and \(\mathbf{v}=(z,w)^{\top}\). The first column is a unit vector, so \(x^2+y^2=1\). A perpendicular unit vector can be obtained by turning \(\mathbf{u}\) to \((-y,x)^{\top}\); the other possibility is its negative, \((y,-x)^{\top}\). The classification theorem states that these are not just examples: they exhaust all possibilities.
Proof. Suppose first that \(\mathbf{Q}\) is orthogonal, and write its columns as \(\mathbf{u}=(x,y)^{\top}\) and \(\mathbf{v}=(z,w)^{\top}\). Then \(\mathbf{u}\cdot\mathbf{u}=1\), \(\mathbf{u}\cdot\mathbf{v}=0\), and \(\mathbf{v}\cdot\mathbf{v}=1\). The vectors \(\mathbf{u}\) and \(\mathbf{r}=(-y,x)^{\top}\) are orthonormal, since \(\mathbf{u}\cdot\mathbf{r}=-xy+xy=0\) and \(\mathbf{r}\cdot\mathbf{r}=x^2+y^2=1\). Express \(\mathbf{v}\) in these perpendicular directions: its coefficient along \(\mathbf{u}\) is \(\mathbf{u}\cdot\mathbf{v}=0\), and its coefficient along \(\mathbf{r}\) is \(t=\mathbf{r}\cdot\mathbf{v}\). Thus \(\mathbf{v}=t\mathbf{r}\). Because \(\mathbf{v}\) and \(\mathbf{r}\) both have length \(1\), \(t^2=1\), so \(t=1\) or \(t=-1\). If \(t=1\), the columns are \((x,y)^{\top}\) and \((-y,x)^{\top}\), giving the first displayed form. Its determinant is \(x^2+y^2=1\). If \(t=-1\), the second column is \((y,-x)^{\top}\), giving the second form; its determinant is \(-x^2-y^2=-1\). Conversely, for \(x^2+y^2=1\), the columns in either displayed form each have squared length \(x^2+y^2=1\), and their dot product is \(0\). Hence their column Gram matrix is \(\mathbf{I}_2\), so each form is orthogonal. The determinant computations just given establish the claimed signs and finish both directions. \(\square
The condition \(x^2+y^2=1\) describes all allowed first columns. For each such first column, choosing its perpendicular column in one direction gives determinant \(1\); choosing the opposite perpendicular column gives determinant \(-1\). Thus the determinant records which of the two possible perpendicular directions was used. It does not determine the matrix by itself: the first column is needed as well.
Worked Classification
Worked Example: Identify the Rotation Form
Classify the matrix \(\mathbf{Q}=\begin{bmatrix}\frac35&-\frac45\\[2pt]\frac45&\frac35\end{bmatrix}\) by deriving its type and checking orthogonality.
Its first column is \((\frac35,\frac45)^{\top}\), whose squared length is \(\frac{9}{25}+\frac{16}{25}=1\). The second column is \((-\frac45,\frac35)^{\top}\). Their dot product is \(\frac35(-\frac45)+\frac45(\frac35)=-\frac{12}{25}+\frac{12}{25}=0\), and the second column also has squared length \(\frac{16}{25}+\frac{9}{25}=1\). Therefore the columns are orthonormal.
Here \(x=\frac35\) and \(y=\frac45\), so the entries match the first form. Directly,
Its determinant is \(\frac35\cdot\frac35-(-\frac45)\cdot\frac45=\frac9{25}+\frac{16}{25}=1\). It is a rotation matrix.
Worked Example: Identify the Reflection Form
Classify \(\mathbf{Q}=\begin{bmatrix}\frac5{13}&\frac{12}{13}\\[2pt]\frac{12}{13}&-\frac5{13}\end{bmatrix}\).
The first column \((\frac5{13},\frac{12}{13})^{\top}\) has squared length \(\frac{25}{169}+\frac{144}{169}=1\). The second column \((\frac{12}{13},-\frac5{13})^{\top}\) has the same squared length, and their dot product is \(\frac{60}{169}-\frac{60}{169}=0\). Thus the columns are orthonormal. The matrix matches the second form with \(x=\frac5{13}\) and \(y=\frac{12}{13}\).
Multiplication verifies orthogonality, including both diagonal and off-diagonal entries:
The determinant is \(-\frac{25}{169}-\frac{144}{169}=-1\). Therefore this is a reflection matrix.
Worked Example: Reconstruct the Second Column
Suppose an orthogonal matrix has first column \((-\frac35,\frac45)^{\top}\) and determinant \(1\). Find its second column and verify the resulting matrix.
For determinant \(1\), the classification theorem requires the second column to be \((-y,x)^{\top}\). Here \(x=-\frac35\) and \(y=\frac45\), so it must be \((-\frac45,-\frac35)^{\top}\). The candidate matrix is
The first column has squared length \(\frac9{25}+\frac{16}{25}=1\); the second has squared length \(\frac{16}{25}+\frac9{25}=1\). Their dot product is \(\frac{12}{25}-\frac{12}{25}=0\), so the columns are orthonormal. Multiplying back gives
Its determinant is \(\frac9{25}+\frac{16}{25}=1\), as required. The other perpendicular choice would give determinant \(-1\), so it would not meet the stated condition.
Why the Determinant Is a Useful Classifier
The two forms provide a quick test once orthogonality is known. The determinant is not merely a label added after the calculation: it distinguishes the two possible arrangements of the columns. For a fixed first column \((x,y)^{\top}\), compare the permitted second columns:
| Determinant | Second column | Matrix form |
|---|---|---|
| \(1\) | \((-y,x)^{\top}\) | \(\begin{bmatrix}x&-y\\y&x\end{bmatrix}\) |
| \(-1\) | \((y,-x)^{\top}\) | \(\begin{bmatrix}x&y\\y&-x\end{bmatrix}\) |
This also gives a practical classification procedure: check that the columns are orthonormal, then calculate the determinant. If the determinant is \(1\), the matrix has the rotation form; if it is \(-1\), it has the reflection form. The form itself is a useful additional check: its parameters must satisfy \(x^2+y^2=1\). A determinant of \(1\) or \(-1\) alone is not enough to prove that an arbitrary matrix is orthogonal.
Common Pitfall: Unit Columns Are Not Enough
A frequent error is to check that both columns have length \(1\) and conclude that the matrix is orthogonal. The columns must also be perpendicular. For instance, consider \(\mathbf{A}=\begin{bmatrix}\frac35&\frac45\\[2pt]\frac45&\frac35\end{bmatrix}\). Each column has squared length \(\frac9{25}+\frac{16}{25}=1\). But their dot product is \(\frac35\cdot\frac45+\frac45\cdot\frac35=\frac{24}{25}\), not \(0\). The wrong conclusion would be “both columns are unit vectors, so \(\mathbf{A}\) is orthogonal.” The correction is to check the off-diagonal entries as well:
So \(\mathbf{A}\) is not orthogonal, even though each column individually has length \(1\). The classification theorem applies only after orthogonality has been established; matching one entry pattern or having unit columns by themselves does not suffice.
The distinction between the two forms prepares a useful way to organize planar orthogonal matrices: first choose a unit first column, then choose one of its two perpendicular unit vectors. The determinant records which choice was made. This is the full classification, while describing the rotation form through angles and analyzing reflections across particular lines are separate geometric questions.
Check Your Understanding
For each problem, use the column conditions and the determinant carefully; where appropriate, verify the result by multiplying \(\mathbf{Q}^{\top}\mathbf{Q}\).
- Write the rotation-form matrix whose first column is \((\frac{8}{17},\frac{15}{17})^{\top}\), and verify its determinant.
- For first column \((\frac{12}{13},-\frac5{13})^{\top}\), write the second column that gives determinant \(-1\).
- Explain why the condition \(\det\mathbf{Q}=1\) does not by itself prove that \(\mathbf{Q}\) is orthogonal.
- Prove directly that the columns of \(\begin{bmatrix}x&y\\y&-x\end{bmatrix}\) are orthonormal whenever \(x^2+y^2=1\).
- Classify \(\begin{bmatrix}0&1\\1&0\end{bmatrix}\) using its form and determinant.
