Conjugate Eigenvalue Pairs For Real Matrices

Why Nonreal Eigenvalues Come in Pairs

A real matrix has real entries, but its eigenvalues need not be real when we allow complex scalars. What the real entries do guarantee is a symmetry: whenever a nonreal number is an eigenvalue, its complex conjugate is also an eigenvalue. The characteristic polynomial makes this symmetry precise. Its coefficients are real, so conjugating an input and its polynomial value preserves the equation that the polynomial equals zero.

In this tutorial, a real \(n\times n\) matrix is regarded as acting on \(\mathbb{C}^n\), with the same entries and the usual matrix-vector multiplication. For a complex number \(z=a+bi\), its conjugate is \(\overline{z}=a-bi\). A number is nonreal if it is not equal to its conjugate.

The last equivalence is the determinant criterion for a square matrix: its determinant is zero exactly when its columns are linearly dependent. If \(\det(\mathbf{A}-\lambda\mathbf{I}_n)=0\), column dependence gives a nonzero vector \(\mathbf{v}\) with \((\mathbf{A}-\lambda\mathbf{I}_n)\mathbf{v}=\mathbf{0}\), which is the eigenvalue equation. Conversely, such a nonzero vector makes the columns dependent and the determinant zero. This criterion lets us check eigenvalues by evaluating the characteristic polynomial.

The Conjugation Argument

Proof. Since \(\mathbf{A}\) is real, conjugating each entry leaves it unchanged: \(\overline{\mathbf{A}}=\mathbf{A}\). Also, conjugation commutes with addition and multiplication of complex numbers. The determinant is a sum of signed products of matrix entries, so conjugating a determinant conjugates each entry in those products. Consequently, for any complex matrix \(\mathbf{M}\), \(\overline{\det\mathbf{M}}=\det(\overline{\mathbf{M}})\). Apply this identity to \(\mathbf{M}=\mathbf{A}-\lambda\mathbf{I}_n\):

If \(\lambda\) is an eigenvalue, then \(p_{\mathbf{A}}(\lambda)=0\). The displayed identity gives \(p_{\mathbf{A}}(\overline{\lambda})=\overline{0}=0\), so \(\overline{\lambda}\) is an eigenvalue by the determinant criterion. Finally, if \(\lambda\) is nonreal, then \(\lambda\ne\overline{\lambda}\), so the two eigenvalues are distinct. \(\square\)

The same fact can be read from the coefficients. Expanding the determinant shows that \(p_{\mathbf{A}}\) has real coefficients, because every entry of \(\mathbf{A}\) is real. A polynomial with real coefficients satisfies \(p_{\mathbf{A}}(\overline{\lambda})=\overline{p_{\mathbf{A}}(\lambda)}\). The determinant argument above proves that identity directly, without needing to list the coefficients of the polynomial.

Worked Examples

Why the Multiplicities Match

The pairing also preserves algebraic multiplicity. The algebraic multiplicity of an eigenvalue is its multiplicity as a root of the characteristic polynomial: it is the largest integer \(m\) for which the polynomial has a factor \((\lambda-\lambda_0)^m\), where \(\lambda_0\) is that root.

Proof. The characteristic polynomial has real coefficients. Suppose \(\lambda\) has algebraic multiplicity \(m\), so it factors as \(p_{\mathbf{A}}(z)=(z-\lambda)^m q(z)\), where \(q\) is a polynomial and \(q(\lambda)\ne0\). Conjugate the coefficients in this factorization. Because the coefficients of \(p_{\mathbf{A}}\) are real, conjugating them leaves \(p_{\mathbf{A}}\) unchanged. The conjugated factorization is therefore

where \(\widetilde{q}\) is obtained by conjugating the coefficients of \(q\), and \(\widetilde{q}(\overline{\lambda})=\overline{q(\lambda)}\ne0\). Thus the factor \((z-\overline{\lambda})^m\) occurs, but the remaining factor is nonzero at \(\overline{\lambda}\), so its multiplicity is exactly \(m\). \(\square\)

Equal algebraic multiplicities concern roots of the characteristic polynomial. They do not, by themselves, assert anything about the dimensions of the corresponding eigenspaces. Those are different quantities, and one should not infer an eigenspace dimension from a root multiplicity.

Common Pitfall: Conjugating the Wrong Quantity

A common error is to argue that a real matrix must have only real eigenvalues. The matrix entries being real does not force every root of its characteristic polynomial to be real; it forces nonreal roots to occur with their conjugates. For \(\mathbf{A}=\begin{bmatrix}0&-1\\1&0\end{bmatrix}\), the calculation \(p_{\mathbf{A}}(\lambda)=\lambda^2+1\) gives the nonreal roots \(i\) and \(-i\), so this matrix is a direct counterexample to the mistaken claim.

Another error is to overlook that conjugating the scalar changes the matrix inside the determinant. In general, \(\overline{\mathbf{A}-\lambda\mathbf{I}_n}=\mathbf{A}-\overline{\lambda}\mathbf{I}_n\) for real \(\mathbf{A}\); it is not \(\mathbf{A}-\lambda\mathbf{I}_n\) unless \(\lambda\) is real. For the same \(2\times2\) example, \(p_{\mathbf{A}}(i)=0\), and conjugating this equation gives \(p_{\mathbf{A}}(-i)=0\), not another assertion about \(p_{\mathbf{A}}(i)\). Keeping track of the conjugated input is the key step.

The practical conclusion is precise: real entries do not rule out complex eigenvalues, but they constrain their pattern. Every nonreal eigenvalue has a distinct conjugate partner, and the two roots occur with equal algebraic multiplicity. A real eigenvalue is unchanged by conjugation and therefore does not require a distinct partner.