The Fundamental Theorem Of Algebra And Complex Eigenvalues

Why Complex Scalars Guarantee an Eigenvalue

An eigenvalue is a scalar for which a matrix acts on some nonzero vector by simple scaling. Over the real numbers, a matrix need not have a real eigenvalue. Allowing complex scalars changes the general existence question: every complex square matrix of positive size has at least one complex eigenvalue. The reason combines two facts. Its characteristic polynomial has positive degree, and every nonconstant polynomial with complex coefficients has a complex root.

We work over \(\mathbb{C}\), and let \(n\) be a positive integer. Thus \(\mathbf{A}\in\mathbb{C}^{n\times n}\). A vector \(\mathbf{v}\in\mathbb{C}^n\) is a column. An eigenvalue of \(\mathbf{A}\) is a scalar \(\lambda\in\mathbb{C}\) for which there is a vector \(\mathbf{v}\ne\mathbf{0}\) satisfying \(\mathbf{A}\mathbf{v}=\lambda\mathbf{v}\); such a vector is an eigenvector for \(\lambda\).

We use the Fundamental Theorem of Algebra in the following form: every nonconstant polynomial with complex coefficients has at least one root in \(\mathbb{C}\). This theorem is a result from the theory of complex numbers; its proof is not part of the linear algebra argument here. The key point for applying it is to verify that the characteristic polynomial really is nonconstant.

The Characteristic Polynomial Has Positive Degree

For an \(n\times n\) matrix, the characteristic polynomial has degree exactly \(n\). To see why, the entry in row \(i\), column \(j\) of \(\mathbf{A}-\lambda\mathbf{I}_n\) is \(a_{ij}-\lambda\delta_{ij}\), where \(\delta_{ij}=1\) if \(i=j\) and \(0\) otherwise. In the determinant expansion, a term of degree \(n\) in \(\lambda\) can only come from choosing the \(-\lambda\) part from each diagonal entry. That contribution is \((-\lambda)^n\). Every other contribution has degree less than \(n\). Therefore the leading coefficient is \((-1)^n\), which is nonzero, and the degree is \(n\).

The positivity assumption \(n\ge 1\) matters: it ensures that this polynomial has positive degree, so the Fundamental Theorem of Algebra applies. The theorem does not say that its roots are distinct, or that every scalar is a root. It guarantees at least one root in \(\mathbb{C}\).

From a Root to an Eigenvalue

The link between a root and an eigenvalue is the determinant criterion for square matrices: for \(\mathbf{B}\in\mathbb{C}^{n\times n}\), \(\det\mathbf{B}=0\) if and only if \(\mathbf{B}\mathbf{x}=\mathbf{0}\) has a nonzero solution \(\mathbf{x}\). Equivalently, a square matrix has determinant zero exactly when its columns are linearly dependent. This standard determinant result applies to square matrices over \(\mathbb{C}\).

Proof. Since \(n\ge1\), the characteristic polynomial \(p_{\mathbf{A}}(\lambda)=\det(\mathbf{A}-\lambda\mathbf{I}_n)\) has degree \(n>0\), as shown above. The Fundamental Theorem of Algebra therefore gives a scalar \(\lambda_0\in\mathbb{C}\) such that \(p_{\mathbf{A}}(\lambda_0)=0\). By the definition of the characteristic polynomial, \(\det(\mathbf{A}-\lambda_0\mathbf{I}_n)=0\). The determinant criterion gives a vector \(\mathbf{v}\ne\mathbf{0}\) with \((\mathbf{A}-\lambda_0\mathbf{I}_n)\mathbf{v}=\mathbf{0}\). Expanding this equation yields \(\mathbf{A}\mathbf{v}-\lambda_0\mathbf{v}=\mathbf{0}\), hence \(\mathbf{A}\mathbf{v}=\lambda_0\mathbf{v}\). Thus \(\lambda_0\) is an eigenvalue of \(\mathbf{A}\). \(\square\)

The argument also proves the useful equivalence that a scalar \(\lambda\) is an eigenvalue of \(\mathbf{A}\) if and only if it is a root of \(p_{\mathbf{A}}\). In one direction, an eigenvector satisfies \((\mathbf{A}-\lambda\mathbf{I}_n)\mathbf{v}=\mathbf{0}\) with \(\mathbf{v}\ne\mathbf{0}\), so the determinant criterion gives \(p_{\mathbf{A}}(\lambda)=0\). In the other direction, if \(p_{\mathbf{A}}(\lambda)=0\), the same criterion produces a nonzero vector satisfying the eigenvector equation.

An Example Where the Root Produces an Eigenvector

What the Theorem Does—and Does Not—Guarantee

The conclusion is an existence statement. It guarantees at least one eigenvalue when the matrix has positive size and its entries are complex. It does not promise that an eigenvalue is real, that the roots are all different, or that the matrix can be diagonalized. Those stronger claims require additional hypotheses or further analysis. In particular, knowing that the characteristic polynomial has a root gives a scalar \(\lambda\); the determinant criterion then gives a nonzero vector, but it does not by itself describe all eigenvalues or all eigenvectors.

The distinction between roots counted with multiplicity and distinct roots is important. A polynomial may have positive degree while all its roots coincide. For instance, the identity matrix of size \(3\) has characteristic polynomial \((1-\lambda)^3\), not a polynomial with three distinct roots.

Common Pitfall: Confusing Existence With a Complete Description

A common incorrect argument is: “The characteristic polynomial has degree \(n\), so it has \(n\) distinct roots, and therefore the matrix has \(n\) distinct eigenvalues.” The conclusion does not follow. The Fundamental Theorem of Algebra guarantees a complex root for a nonconstant polynomial; it does not guarantee distinct roots. The identity-matrix calculation above corrects this: \((1-\lambda)^3\) has degree \(3\), but its only distinct root is \(1\).

A second error is to stop after finding a root and treat that alone as a displayed eigenpair. A root gives \(\det(\mathbf{A}-\lambda\mathbf{I}_n)=0\); the determinant criterion then guarantees a nonzero solution to \((\mathbf{A}-\lambda\mathbf{I}_n)\mathbf{v}=\mathbf{0}\). The vector must be nonzero to meet the definition of eigenvector. In a proof of existence, the criterion guarantees such a vector even when we do not calculate it explicitly. In a worked computation, checking \(\mathbf{A}\mathbf{v}=\lambda\mathbf{v}\) confirms the proposed pair.

There is also a size condition: the argument uses a square matrix with \(n\ge1\). It does not apply to a rectangular matrix, whose characteristic polynomial is not defined by \(\det(\mathbf{A}-\lambda\mathbf{I})\), or to a zero-size matrix, whose characteristic polynomial has no positive degree.

Why This Result Matters

The Fundamental Theorem of Algebra gives complex scalars a central role in eigenvalue theory. For a matrix of any positive size, it ensures that the characteristic equation has at least one solution once the scalars are allowed to be complex. The proof is short, but each link matters: positive size makes the characteristic polynomial nonconstant; the Fundamental Theorem gives a root; and the determinant criterion turns that root into a nonzero eigenvector. This establishes existence without assuming that the matrix is diagonalizable, and without requiring us to solve the characteristic equation explicitly.