Equivalent Tests for Orthogonal Matrices

Four Ways to Recognize an Orthogonal Matrix

For a square real matrix, several natural checks express the same underlying condition. One can check the columns, check the rows, multiply by the transpose on either side, or compare the transpose with the inverse. The equivalence is useful because the most convenient calculation depends on how a matrix is presented. The square-matrix hypothesis matters: the two matrix products have different dimensions for a rectangular matrix, and one identity alone need not imply the other.

Throughout, let \(\mathbf{Q}\in\mathbb{R}^{n\times n}\), where \(n\in\mathbb{N}\). The columns and rows are vectors in \(\mathbb{R}^n\). A list of vectors is orthonormal if each vector has length \(1\) and distinct vectors have dot product \(0\). Equivalently, for vectors \(\mathbf{u}_1,\dots,\mathbf{u}_n\), orthonormality means \(\mathbf{u}_i\cdot\mathbf{u}_j=1\) when \(i=j\), and \(0\) when \(i\ne j\).

What the Matrix Products Record

Write the columns of \(\mathbf{Q}\) as \(\mathbf{q}_1,\dots,\mathbf{q}_n\). The entry in row \(i\), column \(j\) of \(\mathbf{Q}^{\top}\mathbf{Q}\) is the dot product \(\mathbf{q}_i\cdot\mathbf{q}_j\). Thus the diagonal entries record the squared lengths of the columns, and the off-diagonal entries record their pairwise dot products. Likewise, the entry in row \(i\), column \(j\) of \(\mathbf{Q}\mathbf{Q}^{\top}\) is the dot product of rows \(i\) and \(j\) of \(\mathbf{Q}\).

These entry formulas prove the direct tests: the columns are orthonormal exactly when \(\mathbf{Q}^{\top}\mathbf{Q}=\mathbf{I}_n\), and the rows are orthonormal exactly when \(\mathbf{Q}\mathbf{Q}^{\top}=\mathbf{I}_n\). The remaining point is to show why, for a square matrix, either one-sided identity forces the other one as well.

Why One Identity Is Enough for a Square Matrix

We use a basic consequence of row reduction. If the columns of an \(n\times n\) matrix are independent, then each column is a pivot column. There are \(n\) pivots in \(n\) rows, so the reduced row echelon form is \(\mathbf{I}_n\). In particular, row reduction shows that \(\mathbf{A}\mathbf{x}=\mathbf{b}\) has a solution for every \(\mathbf{b}\in\mathbb{R}^n\). This argument is specific to square matrices.

Proof. By the entry formulas, condition (1) is equivalent to \(\mathbf{Q}^{\top}\mathbf{Q}=\mathbf{I}_n\), and condition (2) is equivalent to \(\mathbf{Q}\mathbf{Q}^{\top}=\mathbf{I}_n\). It remains to show that either one-sided identity implies both.

Suppose first that \(\mathbf{Q}^{\top}\mathbf{Q}=\mathbf{I}_n\). If \(\mathbf{Q}\mathbf{x}=\mathbf{0}\), then

So the columns of \(\mathbf{Q}\) are independent. By the row-reduction fact above, \(\mathbf{Q}\mathbf{x}=\mathbf{b}\) has a unique solution for every \(\mathbf{b}\in\mathbb{R}^n\); hence \(\mathbf{Q}\) is invertible. Since \(\mathbf{Q}^{\top}\mathbf{Q}=\mathbf{I}_n\), \(\mathbf{Q}^{\top}\) is a left inverse. The inverse of an invertible matrix is its unique left inverse: if \(\mathbf{L}\mathbf{Q}=\mathbf{I}_n\), then \(\mathbf{L}=\mathbf{L}(\mathbf{Q}\mathbf{Q}^{-1})=(\mathbf{L}\mathbf{Q})\mathbf{Q}^{-1}=\mathbf{Q}^{-1}\). Thus \(\mathbf{Q}^{\top}=\mathbf{Q}^{-1}\), and consequently \(\mathbf{Q}\mathbf{Q}^{\top}=\mathbf{I}_n\).

Now suppose instead that \(\mathbf{Q}\mathbf{Q}^{\top}=\mathbf{I}_n\). Applying the same argument to the square matrix \(\mathbf{Q}^{\top}\) gives \((\mathbf{Q}^{\top})^{-1}=\mathbf{Q}\). Taking inverses yields \(\mathbf{Q}^{-1}=\mathbf{Q}^{\top}\), and so \(\mathbf{Q}^{\top}\mathbf{Q}=\mathbf{I}_n\). This proves that either one-sided identity implies condition (3), and also establishes condition (4). Finally, if condition (4) holds, the definition of an inverse gives both \(\mathbf{Q}^{-1}\mathbf{Q}=\mathbf{I}_n\) and \(\mathbf{Q}\mathbf{Q}^{-1}=\mathbf{I}_n\). Substituting \(\mathbf{Q}^{\top}\) for \(\mathbf{Q}^{-1}\) gives condition (3). All four conditions are therefore equivalent. \(\square\)

Using the Row Test and the Inverse Test

The equivalence theorem lets us choose the calculation that is simplest. If the rows have a visibly simple form, check their dot products or compute \(\mathbf{Q}\mathbf{Q}^{\top}\). If the columns are given in a problem, compute \(\mathbf{Q}^{\top}\mathbf{Q}\). If an inverse is already known, verify that it equals the transpose. In every case, the matrix must be square for the theorem to connect the tests.

Common Pitfall: Unit Columns Need Not Be Orthonormal

Checking only that every column has length \(1\) is not enough. For example, consider \(\mathbf{A}=\begin{bmatrix}1&1\\0&0\end{bmatrix}\). Both columns have length \(1\), but their dot product is \(1\), not \(0\). The incorrect calculation “both columns are unit vectors, so the matrix is orthogonal” omits the pairwise orthogonality condition.

The correction is to check all column dot products, or equivalently to compute the entire matrix \(\mathbf{A}^{\top}\mathbf{A}\). Its off-diagonal entries here are \(1\), revealing that the columns are not orthogonal. The rows fail the test as well: the first row has squared length \(2\), and the second row has squared length \(0\). Thus none of the equivalent orthogonal-matrix conditions holds.

There is also a shape issue to keep in view. For a rectangular matrix, \(\mathbf{Q}^{\top}\mathbf{Q}\) and \(\mathbf{Q}\mathbf{Q}^{\top}\) have different sizes, and a one-sided identity is not the square-matrix equivalence proved here. Check the dimensions and the stated hypotheses before using the theorem.