Real Normal Matrices That Are Not Symmetric

Normal Does Not Mean Symmetric

A real matrix can preserve lengths and still fail to be symmetric. A planar rotation makes this distinction visible: rotating the plane preserves the lengths and angles between vectors, but a nontrivial rotation usually does not equal its transpose. Its eigenvalues also reveal a difference. Although the matrix has real entries, its eigenvalues are generally nonreal when vectors and eigenvalues are considered over \(\mathbb{C}\).

For a real square matrix \(\mathbf{A}\), normality means that \(\mathbf{A}^{\top}\mathbf{A}=\mathbf{A}\mathbf{A}^{\top}\). Symmetry means the stronger-looking entrywise condition \(\mathbf{A}^{\top}=\mathbf{A}\). A real matrix is orthogonal if \(\mathbf{A}^{\top}\mathbf{A}=\mathbf{I}\); every orthogonal matrix is normal, since in that case both products equal \(\mathbf{I}\). The calculations below show that rotations are orthogonal, hence normal, but are not necessarily symmetric.

Write \(c=\cos\theta\) and \(s=\sin\theta\). The rotation by angle \(\theta\) is represented in the standard coordinates of \(\mathbb{R}^2\) by

Checking Normality and Symmetry

To check normality, compute both products rather than infer normality from the geometric description alone. The transpose and products are

Multiplying in the other order gives

Both calculations use \(c^2+s^2=1\). Therefore every planar rotation is orthogonal and normal. For symmetry, compare the off-diagonal entries: the transpose has entries \(s\) and \(-s\) where the original has \(-s\) and \(s\). Equality requires \(-s=s\), which over \(\mathbb{R}\) is equivalent to \(s=0\). Thus a rotation is symmetric exactly when \(\sin\theta=0\); otherwise it is a real normal matrix that is not symmetric.

Proof. The product calculations above show that \(\mathbf{R}_{\theta}^{\top}\mathbf{R}_{\theta}=\mathbf{R}_{\theta}\mathbf{R}_{\theta}^{\top}=\mathbf{I}_2\), so it is normal. Comparing the two off-diagonal entries shows that \(\mathbf{R}_{\theta}^{\top}=\mathbf{R}_{\theta}\) if and only if \(-s=s\), or \(s=0\). To establish the eigenvalue statement, use the two nonzero complex vectors \(\mathbf{z}_{+}=(1,-i)^{\top}\) and \(\mathbf{z}_{-}=(1,i)^{\top}\). Direct multiplication gives \(\mathbf{R}_{\theta}\mathbf{z}_{+}=(c+is,\;s-ic)^{\top}=(c+is)\mathbf{z}_{+}\) and \(\mathbf{R}_{\theta}\mathbf{z}_{-}=(c-is,\;s+ic)^{\top}=(c-is)\mathbf{z}_{-}\). Thus the two stated scalars are eigenvalues. They are nonreal precisely when \(s\ne0\), the same condition under which the matrix is not symmetric. \(\square\)

The same eigenvalues appear in the characteristic polynomial, using the convention \(p_{\mathbf{A}}(\lambda)=\det(\mathbf{A}-\lambda\mathbf{I})\). For the rotation, expansion gives

Its factorization is \((\lambda-(c+is))(\lambda-(c-is))\). This algebra agrees with the eigenvectors checked in the proof. When \(s\ne0\), the two roots form a nonreal conjugate pair; when \(s=0\), they coincide at a real value.

Worked Rotations

Why Real Symmetric Matrices Behave Differently

The contrast is not that symmetric matrices are normal while rotations are not: both classes are normal. The distinction is that a real symmetric matrix cannot have a nonreal eigenvalue, whereas a nonsymmetric planar rotation does. Here is a proof that applies to real symmetric matrices of any finite size.

Proof. Suppose \(\lambda\in\mathbb{C}\) is an eigenvalue and \(\mathbf{z}\in\mathbb{C}^n\) is a corresponding nonzero eigenvector, so \(\mathbf{A}\mathbf{z}=\lambda\mathbf{z}\). Write \(\mathbf{z}^{*}\) for the conjugate transpose of \(\mathbf{z}\). Since \(\mathbf{A}\) is real symmetric, \(\mathbf{A}^{*}=\mathbf{A}\). Consequently, \(\overline{\mathbf{z}^{*}\mathbf{A}\mathbf{z}}=\mathbf{z}^{*}\mathbf{A}^{*}\mathbf{z}=\mathbf{z}^{*}\mathbf{A}\mathbf{z}\), so the scalar \(\mathbf{z}^{*}\mathbf{A}\mathbf{z}\) is real. The eigenvector equation also gives \(\mathbf{z}^{*}\mathbf{A}\mathbf{z}=\lambda\mathbf{z}^{*}\mathbf{z}\). Now \(\mathbf{z}^{*}\mathbf{z}=\sum_{j=1}^{n}|z_j|^2\) is a positive real number because \(\mathbf{z}\ne\mathbf{0}\). Thus \(\lambda=(\mathbf{z}^{*}\mathbf{A}\mathbf{z})/(\mathbf{z}^{*}\mathbf{z})\) is real. \(\square\)

This result explains why the quarter-turn cannot be symmetric: its eigenvalues \(i\) and \(-i\) are nonreal. More generally, whenever \(\sin\theta\ne0\), the rotation theorem gives nonreal eigenvalues, which rules out symmetry by the result just proved. Conversely, when \(\sin\theta=0\), the rotation matrix is symmetric, and its eigenvalues are real.

Common Pitfall: Real Entries Do Not Force Real Eigenvalues

A common error is to conclude that a matrix with real entries must have real eigenvalues. The quarter-turn is a concrete counterexample. A mistaken factorization might turn its characteristic polynomial \(\lambda^2+1\) into \((\lambda+1)^2\), leading to the incorrect eigenvalue \(-1\). But \((\lambda+1)^2=\lambda^2+2\lambda+1\), which is not \(\lambda^2+1\). The correct roots are \(i\) and \(-i\), as the explicit eigenvector checks show.

It is also important to distinguish normality from symmetry. The equality \(\mathbf{A}^{\top}\mathbf{A}=\mathbf{A}\mathbf{A}^{\top}\) compares two products; it does not assert \(\mathbf{A}^{\top}=\mathbf{A}\). For a rotation, both products equal \(\mathbf{I}_2\), while the transpose changes the signs of the off-diagonal entries. Confusing these conditions would incorrectly classify every rotation as symmetric.

The practical lesson is to check each property by its own definition. For a candidate real normal matrix, compute \(\mathbf{A}^{\top}\mathbf{A}\) and \(\mathbf{A}\mathbf{A}^{\top}\) to test normality, compare \(\mathbf{A}\) with \(\mathbf{A}^{\top}\) to test symmetry, and allow complex vectors when investigating eigenvalues. These checks separate three ideas that can look similar when one first encounters them.