Real Eigenvalues of Symmetric Matrices

The Question Behind Real Eigenvalues

A real matrix acts on real vectors, but its eigenvalue equation can also be considered over \(\mathbb{C}\). In that setting, a vector may have complex entries and its eigenvalue may initially appear to be complex. For a real symmetric matrix, however, a nonreal eigenvalue is impossible. The proof rests on a scalar expression built from a complex eigenvector: its value must be real because of symmetry, while the eigenvalue equation also expresses it as the eigenvalue times a positive real number.

Let \(\mathbf{A}\in\mathbb{R}^{n\times n}\). When we consider complex vectors, we regard the same entries of \(\mathbf{A}\) as complex numbers and let \(\mathbf{A}\) act on \(\mathbb{C}^n\). A complex eigenvalue \(\lambda\) means that there is a nonzero vector \(\mathbf{x}\in\mathbb{C}^n\) such that \(\mathbf{A}\mathbf{x}=\lambda\mathbf{x}\). The eigenvector need not have real entries. The transpose criterion for symmetry is \(\mathbf{A}^{\top}=\mathbf{A}\), and because \(\mathbf{A}\) is real, this also gives \(\mathbf{A}^{*}=\mathbf{A}\).

The Conjugate-Transpose Identity

The key calculation concerns the scalar \(\mathbf{x}^{*}\mathbf{A}\mathbf{x}\). It is a \(1\times1\) matrix, which we treat as a scalar. Taking the conjugate of a scalar reverses the order of factors when expressed as a conjugate transpose. In particular, for any complex square matrix \(\mathbf{A}\) and any \(\mathbf{x}\in\mathbb{C}^n\),

The first equality holds because the conjugate transpose of a scalar is its complex conjugate. For the second, reverse the factor order and take conjugate transposes: \((\mathbf{x}^{*})^{*}=\mathbf{x}\), and the leftmost resulting factor is \(\mathbf{x}^{*}\). If \(\mathbf{A}\) is real symmetric, then \(\mathbf{A}^{*}=\mathbf{A}^{\top}=\mathbf{A}\), so this identity says that \(\mathbf{x}^{*}\mathbf{A}\mathbf{x}\) equals its own complex conjugate.

Proof. Since \(\mathbf{A}\) is real symmetric, \(\mathbf{A}^{*}=\mathbf{A}\). Applying the conjugate-transpose identity gives \(\overline{\mathbf{x}^{*}\mathbf{A}\mathbf{x}}=\mathbf{x}^{*}\mathbf{A}^{*}\mathbf{x}=\mathbf{x}^{*}\mathbf{A}\mathbf{x}\). A complex scalar equals its conjugate exactly when it is real, so \(\mathbf{x}^{*}\mathbf{A}\mathbf{x}\in\mathbb{R}\). \(\square\)

Why Every Complex Eigenvalue Is Real

The result about \(\mathbf{x}^{*}\mathbf{A}\mathbf{x}\) applies to any complex vector. Now suppose the vector is an eigenvector. Multiplying the eigenvalue equation on the left by \(\mathbf{x}^{*}\) connects that real scalar to the eigenvalue.

Proof. Assume \(\lambda\in\mathbb{C}\) is an eigenvalue of \(\mathbf{A}\). By definition, there is a vector \(\mathbf{x}\in\mathbb{C}^n\), with \(\mathbf{x}\ne\mathbf{0}\), such that \(\mathbf{A}\mathbf{x}=\lambda\mathbf{x}\). Multiply both sides on the left by \(\mathbf{x}^{*}\):

By the result just proved, the left side is real. Also, \(\mathbf{x}^{*}\mathbf{x}=\sum_{j=1}^{n}|x_j|^2\) is real and strictly positive because \(\mathbf{x}\ne\mathbf{0}\). Dividing the equation by this positive real scalar gives

The numerator and denominator are real, and the denominator is nonzero. Therefore \(\lambda\in\mathbb{R}\). This proves the claim for every complex eigenvalue, without assuming that the eigenvector itself is real. \(\square\)

Checking the Argument on Eigenvectors

What the Quotient Does—and Does Not—Show

For a nonzero vector \(\mathbf{x}\), the quotient \((\mathbf{x}^{*}\mathbf{A}\mathbf{x})/(\mathbf{x}^{*}\mathbf{x})\) is defined because its denominator is positive. If \(\mathbf{A}\) is real symmetric, the numerator is real, so the quotient is real. But a real value of this quotient does not by itself make \(\mathbf{x}\) an eigenvector. The eigenvalue proof uses both facts: the quotient is real, and the eigenvalue equation makes the quotient equal to \(\lambda\).

A Real Eigenvalue Has a Real Eigenvector

The theorem does not say that complex eigenvectors must be real. It does imply a related fact for real symmetric matrices: if a real eigenvalue has a complex eigenvector, then there is also a nonzero real eigenvector for that eigenvalue. This follows by separating the real and imaginary parts of the vector.

Proof. Write \(\mathbf{x}=\mathbf{u}+i\mathbf{v}\), where \(\mathbf{u},\mathbf{v}\in\mathbb{R}^n\). Because \(\mathbf{A}\) and \(\lambda\) are real, the eigenvalue equation becomes \(\mathbf{A}\mathbf{u}+i\mathbf{A}\mathbf{v}=\lambda\mathbf{u}+i\lambda\mathbf{v}\). Equality of the real and imaginary parts gives \(\mathbf{A}\mathbf{u}=\lambda\mathbf{u}\) and \(\mathbf{A}\mathbf{v}=\lambda\mathbf{v}\). Since \(\mathbf{x}\ne\mathbf{0}\), at least one of \(\mathbf{u}\) and \(\mathbf{v}\) is nonzero. That nonzero vector is real and is an eigenvector for \(\lambda\). \(\square\)

Common Pitfall: Using Transpose Instead

For complex vectors, transpose and conjugate transpose are different. Replacing \(\mathbf{x}^{*}\) by \(\mathbf{x}^{\top}\) invalidates the key realness argument: symmetry gives \(\mathbf{A}^{\top}=\mathbf{A}\), but it does not imply that \(\mathbf{x}^{\top}\mathbf{A}\mathbf{x}\) is real when \(\mathbf{x}\) has complex entries.

For a concrete calculation, take the real symmetric matrix \(\mathbf{A}=\begin{bmatrix}1&0\\0&2\end{bmatrix}\) and the complex vector \(\mathbf{x}=\begin{bmatrix}1+i\\1\end{bmatrix}\). The incorrect transpose calculation gives

which is not real. The correction is to conjugate as well as transpose: \(\mathbf{x}^{*}=\begin{bmatrix}1-i&1\end{bmatrix}\). Then

which is real, as the conjugate-transpose identity guarantees. Also, the strict positivity of \(\mathbf{x}^{*}\mathbf{x}\) depends on \(\mathbf{x}\ne\mathbf{0}\); that hypothesis is essential when dividing to obtain \(\lambda\).