Characteristic Polynomials Of Two-By-Two Matrices

From Matrix Entries to a Polynomial

For a two-by-two matrix, the characteristic polynomial can be calculated directly from a determinant. The calculation is short, but its signs matter: the diagonal entries become expressions involving \(-\lambda\), while the off-diagonal entries do not change. Writing out this small determinant also reveals exactly how the matrix’s trace and determinant appear in the polynomial.

Throughout, let \(\mathbf{A}\in\mathbb{R}^{2\times2}\), and use the convention \(p_{\mathbf{A}}(\lambda)=\det(\mathbf{A}-\lambda\mathbf{I}_2)\). Here \(\lambda\) is an indeterminate, so \(p_{\mathbf{A}}\) is a polynomial with real coefficients. For \(\mathbf{A}=\begin{bmatrix}a&b\\c&d\end{bmatrix}\), the trace is \(\operatorname{tr}\mathbf{A}=a+d\), and the determinant is \(\det\mathbf{A}=ad-bc\).

The identity matrix in this definition is \(\mathbf{I}_2=\begin{bmatrix}1&0\\0&1\end{bmatrix}\). Subtracting \(\lambda\mathbf{I}_2\) subtracts \(\lambda\) from each diagonal entry; it does not change either off-diagonal entry. The determinant rule for a two-by-two matrix, \(\det\begin{bmatrix}x&y\\z&w\end{bmatrix}=xw-yz\), then gives the formula below.

Proof. By definition and the two-by-two determinant rule,

Substituting \(\operatorname{tr}\mathbf{A}=a+d\) and \(\det\mathbf{A}=ad-bc\) gives the stated expression. Each equality is an expansion or a substitution of the definitions, so the formula holds for every real two-by-two matrix. \(\square\)

The formula says that the polynomial is monic: its leading coefficient is \(1\). Its coefficient of \(\lambda\) is the negative of the trace, and its constant term is the determinant. These are exact coefficient statements, not approximations. In particular, the trace and determinant can be read from the polynomial once it has been written in descending powers of \(\lambda\).

Three Calculations and Their Checks

What the Coefficients Determine

The formula also gives a precise comparison rule. For two-by-two matrices, equality of characteristic polynomials is equivalent to equality of both trace and determinant. The equivalence concerns the polynomials themselves; it does not claim that the matrices have the same entries.

Proof. Write \(t_{\mathbf{A}}=\operatorname{tr}\mathbf{A}\), \(d_{\mathbf{A}}=\det\mathbf{A}\), and define \(t_{\mathbf{B}}\) and \(d_{\mathbf{B}}\) in the same way. If \(t_{\mathbf{A}}=t_{\mathbf{B}}\) and \(d_{\mathbf{A}}=d_{\mathbf{B}}\), the formula already proved gives \(p_{\mathbf{A}}(\lambda)=\lambda^2-t_{\mathbf{A}}\lambda+d_{\mathbf{A}}=\lambda^2-t_{\mathbf{B}}\lambda+d_{\mathbf{B}}=p_{\mathbf{B}}(\lambda)\). Conversely, suppose \(p_{\mathbf{A}}=p_{\mathbf{B}}\). Then \(\lambda^2-t_{\mathbf{A}}\lambda+d_{\mathbf{A}}=\lambda^2-t_{\mathbf{B}}\lambda+d_{\mathbf{B}}\) as polynomials. Equal polynomials have equal coefficients of each power of \(\lambda\). Comparing the coefficients of \(\lambda\) gives \(-t_{\mathbf{A}}=-t_{\mathbf{B}}\), so \(t_{\mathbf{A}}=t_{\mathbf{B}}\); comparing constant terms gives \(d_{\mathbf{A}}=d_{\mathbf{B}}\). This proves both directions. \(\square\)

Common Sign Errors

Common mistake: A frequent incorrect calculation is to use \(\lambda^2+(\operatorname{tr}\mathbf{A})\lambda+\det\mathbf{A}\), putting a plus sign before the trace term. For example, for \(\mathbf{A}=\begin{bmatrix}2&3\\-1&4\end{bmatrix}\), this wrong rule would give \(\lambda^2+6\lambda+11\). But the definition gives \((2-\lambda)(4-\lambda)-3(-1)=\lambda^2-6\lambda+11\). The correction is to expand the two diagonal factors: the linear terms are \(-2\lambda\) and \(-4\lambda\), so their sum is \(-6\lambda\). In general, this is why the coefficient is \(-\operatorname{tr}\mathbf{A}\), not \(+\operatorname{tr}\mathbf{A}\).

A second source of errors is forgetting that the determinant subtracts the product of the off-diagonal entries. For \(\mathbf{B}\) above, the off-diagonal product is \((-2)(3)=-6\), and the determinant contribution is \(-(-6)=+6\). Replacing it by \(-6\) would give the wrong constant term. Keeping the expression \(ad-bc\) intact until the signs of \(b\) and \(c\) have been substituted helps prevent this error.

Use the formula only after confirming the matrix is two-by-two. It follows from the two-by-two determinant rule; a matrix of a different size requires a determinant calculation appropriate to that size. For a two-by-two matrix, a reliable routine is to identify \(a,b,c,d\), compute \(a+d\) and \(ad-bc\), and assemble \(\lambda^2-(a+d)\lambda+(ad-bc)\). If there is any doubt, expand \(\det(\mathbf{A}-\lambda\mathbf{I}_2)\) independently and compare coefficients.

Why This Form Is Useful

Writing the characteristic polynomial in terms of trace and determinant separates two kinds of information. The trace supplies the coefficient of the linear term, while the determinant supplies the constant term. The worked comparison shows that matching traces alone does not ensure matching characteristic polynomials: the determinants must match as well. Conversely, for two-by-two matrices, matching both quantities is sufficient, as the proved equivalence shows.

Earlier material on characteristic polynomials establishes their connection with eigenvalues. This calculation provides an efficient way to form the polynomial in dimension two before using that connection. The task here is to compute and check the polynomial, not to solve it: first make the determinant expansion correct, then identify its coefficients without changing the sign convention.