Symmetric Matrices as Real Normal Matrices

Symmetry Gives Normality

For a real square matrix, symmetry is a condition on its entries: the matrix equals its transpose. Normality is a different condition, expressed by comparing two matrix products. A symmetric matrix satisfies the normality condition for a direct reason: replacing its transpose by the matrix itself makes both products the same square. The implication is useful, but its direction matters. Showing that every symmetric matrix is normal does not, by itself, show that every normal matrix is symmetric.

All matrices and scalars in this tutorial are real. For a matrix \(\mathbf{A}\in\mathbb{R}^{n\times n}\), its transpose \(\mathbf{A}^{\top}\) is obtained by interchanging rows and columns. The two definitions below concern square matrices, since the products in the normality condition must be defined and have the same size.

These definitions ask different questions. Symmetry requires equality of two matrices, while normality requires equality of two products. In particular, normality does not mean that either product equals \(\mathbf{A}\), or that \(\mathbf{A}\) equals its transpose. To test normality directly, compute both products in the stated order; matrix multiplication is not generally commutative.

Proof. Let \(\mathbf{A}\in\mathbb{R}^{n\times n}\) be symmetric. By definition, \(\mathbf{A}^{\top}=\mathbf{A}\). Substitute this equality into each side of the normality condition: \(\mathbf{A}^{\top}\mathbf{A}=\mathbf{A}\mathbf{A}\) and \(\mathbf{A}\mathbf{A}^{\top}=\mathbf{A}\mathbf{A}\). Both products equal \(\mathbf{A}^2\), so \(\mathbf{A}^{\top}\mathbf{A}=\mathbf{A}\mathbf{A}^{\top}\). Therefore \(\mathbf{A}\) is normal. \(\square\)

The proof uses only the definition of symmetry and the definition of matrix multiplication. It does not require that \(\mathbf{A}\) be invertible, have distinct eigenvalues, or have any particular entries. The size \(n\) can be any positive integer.

Checking the Products in Examples

Orthogonal Coordinates Preserve Normality

The normality condition is also compatible with an orthogonal change of coordinates. A real square matrix \(\mathbf{Q}\) is orthogonal if \(\mathbf{Q}^{\top}\mathbf{Q}=\mathbf{Q}\mathbf{Q}^{\top}=\mathbf{I}\), where \(\mathbf{I}\) is the identity matrix of the appropriate size. The matrices \(\mathbf{A}\) and \(\mathbf{Q}^{\top}\mathbf{A}\mathbf{Q}\) represent the same linear transformation in orthogonally related coordinates. The following result shows that changing coordinates in this way does not alter whether the matrix is normal.

Proof. Set \(\mathbf{B}=\mathbf{Q}^{\top}\mathbf{A}\mathbf{Q}\). The transpose rule for products gives \(\mathbf{B}^{\top}=\mathbf{Q}^{\top}\mathbf{A}^{\top}\mathbf{Q}\). Using \(\mathbf{Q}\mathbf{Q}^{\top}=\mathbf{I}\), compute both products: \(\mathbf{B}^{\top}\mathbf{B}=\mathbf{Q}^{\top}\mathbf{A}^{\top}\mathbf{Q}\mathbf{Q}^{\top}\mathbf{A}\mathbf{Q}=\mathbf{Q}^{\top}\mathbf{A}^{\top}\mathbf{A}\mathbf{Q}\), and \(\mathbf{B}\mathbf{B}^{\top}=\mathbf{Q}^{\top}\mathbf{A}\mathbf{Q}\mathbf{Q}^{\top}\mathbf{A}^{\top}\mathbf{Q}=\mathbf{Q}^{\top}\mathbf{A}\mathbf{A}^{\top}\mathbf{Q}\). If \(\mathbf{A}\) is normal, then \(\mathbf{A}^{\top}\mathbf{A}=\mathbf{A}\mathbf{A}^{\top}\), so these expressions are equal and \(\mathbf{B}\) is normal. Conversely, if \(\mathbf{B}\) is normal, the expressions are equal. Multiply that equality on the left by \(\mathbf{Q}\) and on the right by \(\mathbf{Q}^{\top}\). Since \(\mathbf{Q}\mathbf{Q}^{\top}=\mathbf{I}\), the result is \(\mathbf{A}^{\top}\mathbf{A}=\mathbf{A}\mathbf{A}^{\top}\), so \(\mathbf{A}\) is normal. Both directions follow. \(\square\)

This result is consistent with orthogonal diagonalization of real symmetric matrices, established in the earlier material on the Spectral Theorem. A diagonal real matrix is symmetric and hence normal; orthogonal changes of coordinates preserve normality by the theorem just proved. The direct proof that symmetry implies normality, however, does not depend on diagonalizing the matrix.

Why the Converse Needs Its Own Argument

Let \(S\) denote the collection of real symmetric square matrices and \(N\) the collection of real normal square matrices, considering matrices of each fixed size. The first theorem proves the inclusion \(S\subseteq N\): every member of \(S\) belongs to \(N\). It does not prove \(N\subseteq S\). In logical terms, the statement “if a matrix is symmetric, then it is normal” has the form \(P\Rightarrow Q\). It does not entail the reversed statement \(Q\Rightarrow P\).

This distinction is not a technicality in the proof. The proof begins by assuming \(\mathbf{A}^{\top}=\mathbf{A}\), then uses that equality to replace \(\mathbf{A}^{\top}\) in the two products. A normal matrix is only known to satisfy \(\mathbf{A}^{\top}\mathbf{A}=\mathbf{A}\mathbf{A}^{\top}\), which is an equality of products. That condition does not supply the assumption \(\mathbf{A}^{\top}=\mathbf{A}\) used in the proof. To decide whether the converse holds, one must examine it separately; it cannot be obtained by reversing the steps of a one-way implication.

A common incorrect argument is: “Normality says \(\mathbf{A}^{\top}\mathbf{A}=\mathbf{A}\mathbf{A}^{\top}\); therefore cancel \(\mathbf{A}\) and conclude \(\mathbf{A}^{\top}=\mathbf{A}\).” This is not a valid cancellation. The factors are not arranged as a shared factor on the same side, and the definition of normality does not assume \(\mathbf{A}\) is invertible. Even cancellation with a common left factor can fail for matrices: with \(\mathbf{O}=\begin{bmatrix}0&0\\0&0\end{bmatrix}\), \(\mathbf{I}_2\), and \(\mathbf{O}\), one has \(\mathbf{O}\mathbf{I}_2=\mathbf{O}\mathbf{O}=\mathbf{O}\), although \(\mathbf{I}_2\ne\mathbf{O}\). The correction is to keep the product equality in its given order and use a separate proof if a converse is claimed.

Another tempting shortcut is to transpose the normality equation and treat the result as new information. But transposing \(\mathbf{A}^{\top}\mathbf{A}=\mathbf{A}\mathbf{A}^{\top}\) gives \(\mathbf{A}^{\top}\mathbf{A}=\mathbf{A}\mathbf{A}^{\top}\) again, since each product is unchanged by transposition. Repeating the same condition does not turn an equality of products into equality of the factors.