Trace, Determinant, And The Two-By-Two Eigenvalues

What Trace and Determinant Tell Us About the Roots

For a two-by-two matrix, the characteristic polynomial is a quadratic, so its two roots contain all the eigenvalue information that the polynomial records. Its coefficients are also familiar matrix quantities: the trace and determinant. The connection is precise: counting a repeated root twice when necessary, the sum of the two roots is the trace, and their product is the determinant.

We work with a matrix \(\mathbf{A}\in\mathbb{R}^{2\times2}\). Its roots are considered in \(\mathbb{C}\), so the statement includes matrices whose characteristic polynomial has nonreal roots. A root is an eigenvalue, as established by the eigenvalue equation and the definition of the characteristic polynomial. For this tutorial, the essential task is to relate those roots to two quantities computed directly from the entries of \(\mathbf{A}\).

Expanding the Characteristic Polynomial

Writing out \(\mathbf{A}-\lambda\mathbf{I}_2\) gives \(\begin{bmatrix}a-\lambda&b\\c&d-\lambda\end{bmatrix}\). Its determinant is \((a-\lambda)(d-\lambda)-bc\). Expanding each term shows exactly where the trace and determinant appear:

The leading coefficient is \(1\), so this is a monic quadratic. The minus sign on the trace term matters: it is the negative of the coefficient of \(\lambda\) that gives the sum of the roots. The constant term is the determinant.

Proof. The expansion above gives \(p_{\mathbf{A}}(\lambda)=\lambda^2-(\operatorname{tr}\mathbf{A})\lambda+\det\mathbf{A}\). Since the polynomial is quadratic, its roots counted with multiplicity give the factorization \(p_{\mathbf{A}}(\lambda)=(\lambda-\lambda_1)(\lambda-\lambda_2)\). Expanding that factorization gives \(p_{\mathbf{A}}(\lambda)=\lambda^2-(\lambda_1+\lambda_2)\lambda+\lambda_1\lambda_2\). These are equal polynomials. Equality of the coefficients of \(\lambda\) yields \(-\operatorname{tr}\mathbf{A}=-(\lambda_1+\lambda_2)\), and equality of the constant terms yields \(\det\mathbf{A}=\lambda_1\lambda_2\). Thus both identities hold, including when the roots coincide. \(\square\)

The quadratic formula also makes the roots explicit. If \(t=\operatorname{tr}\mathbf{A}\) and \(d=\det\mathbf{A}\), then the characteristic polynomial is \(\lambda^2-t\lambda+d\), and completing the square gives \((\lambda-\tfrac{t}{2})^2=\tfrac{t^2-4d}{4}\). Thus the roots are \(\tfrac{t}{2}\pm\tfrac{\sqrt{t^2-4d}}{2}\), with the square root taken in \(\mathbb{C}\) when needed. This expression is consistent with the theorem: the opposite square-root terms cancel in the sum, while their product gives the constant term.

Checking the Identities in Examples

What a Repeated Root Says About Trace and Determinant

The sum-and-product identities give a quick test for when the characteristic polynomial has a repeated root. This statement concerns the roots of the polynomial; it does not, by itself, assert how many linearly independent eigenvectors correspond to that root.

Proof. Write \(t=\operatorname{tr}\mathbf{A}\) and \(d=\det\mathbf{A}\), so \(p_{\mathbf{A}}(\lambda)=\lambda^2-t\lambda+d\). If the repeated root is \(r\), then \(p_{\mathbf{A}}(\lambda)=(\lambda-r)^2=\lambda^2-2r\lambda+r^2\). Comparing coefficients gives \(t=2r\) and \(d=r^2\). Therefore \(t^2=4r^2=4d\), and \(r=t/2\). Conversely, suppose \(t^2=4d\). Then $$ p_{\mathbf{A}}(\lambda)=\lambda^2-t\lambda+\frac{t^2}{4} =\left(\lambda-\frac{t}{2}\right)^2. $$ This polynomial has the repeated root \(t/2\). Both directions have been established. \(\square\)

Common Sign and Counting Pitfalls

Do not read the sum directly as the coefficient of \(\lambda\). For example, take \(\mathbf{D}=\begin{bmatrix}2&1\\1&2\end{bmatrix}\). Its trace is \(2+2=4\), and its determinant is \(2\cdot2-1\cdot1=3\). Expanding gives \(p_{\mathbf{D}}(\lambda)=(2-\lambda)^2-1=\lambda^2-4\lambda+3\). A wrong reading would say that the roots sum to \(-4\), using the coefficient of \(\lambda\) without changing its sign. In fact, \(\lambda^2-4\lambda+3=(\lambda-1)(\lambda-3)\), so the roots sum to \(1+3=4\), as required. The coefficient is \(-4\); its negative is the sum.

Do not omit multiplicity. In the repeated-root example, listing the root only once would give sum \(3\) and product \(3\), neither of which matches trace \(6\) and determinant \(9\). The polynomial \((\lambda-3)^2\) has two roots when multiplicities are counted: \(3\) and \(3\).

The characteristic polynomial packages both quantities into its coefficients: its linear coefficient is the negative trace and its constant coefficient is the determinant. This is useful as a consistency check when computing roots, and as a way to recover the trace or determinant from a known pair of eigenvalues. The identities apply whether the roots are distinct or repeated, real or nonreal. For a real matrix with nonreal roots, as in the second example, the roots are complex but their sum and product still match the real trace and determinant.