Why Trace and Determinant Decide the Root Type
For a real \(2\times2\) matrix, the characteristic equation is quadratic. Its coefficients can be read directly from two familiar quantities: the trace, which is the sum of the diagonal entries, and the determinant. The discriminant of that quadratic tells us whether its roots are real or nonreal. This gives a quick test for nonreal eigenvalues without first solving for eigenvectors.
Throughout, let \(\mathbf{A}\in\mathbb{R}^{2\times2}\). We regard its eigenvalues as complex numbers: a scalar \(\lambda\in\mathbb{C}\) is an eigenvalue when \(\mathbf{A}-\lambda\mathbf{I}_2\) is singular, equivalently when \(\det(\mathbf{A}-\lambda\mathbf{I}_2)=0\). The characteristic polynomial convention used here is \(p_{\mathbf{A}}(\lambda)=\det(\mathbf{A}-\lambda\mathbf{I}_2)\).
Expanding the determinant gives the characteristic polynomial in terms of these quantities. In particular, the coefficient of \(\lambda\) is the negative of the trace. Keeping that sign straight matters: it determines the centre of the two roots.
The Discriminant Test
Proof. Write \(T=\operatorname{tr}\mathbf{A}\) and \(D=\det\mathbf{A}\). The characteristic equation is \(\lambda^2-T\lambda+D=0\). Completing the square gives \((2\lambda-T)^2=T^2-4D=\delta\). If \(\delta>0\), the two square roots of \(\delta\) are the distinct real numbers \(\sqrt{\delta}\) and \(-\sqrt{\delta}\), so the roots are \((T+\sqrt{\delta})/2\) and \((T-\sqrt{\delta})/2\). If \(\delta=0\), the only root is \(T/2\), occurring twice. If \(\delta<0\), the two square roots of \(\delta\) are \(i\sqrt{-\delta}\) and \(-i\sqrt{-\delta}\), so the roots are \((T+i\sqrt{-\delta})/2\) and \((T-i\sqrt{-\delta})/2\). These are nonreal because \(\sqrt{-\delta}>0\), and they are conjugates because \(T\) and \(\sqrt{-\delta}\) are real. Thus the characteristic roots are nonreal exactly when \(\delta<0\). \(\square\)
The formula also shows how to read the pair’s location in the complex plane: its real part is half the trace, and the magnitude of its imaginary part is half the square root of the negative discriminant. In the nonreal case, the determinant must be positive. Indeed, \(\delta<0\) means \(T^2-4D<0\), so \(D>T^2/4\geq0\). A positive determinant alone is not enough to guarantee nonreal eigenvalues; the discriminant test uses both trace and determinant.
Proof. The characteristic polynomial is monic and has these two roots, so it factors as \((\lambda-(\alpha+i\beta))(\lambda-(\alpha-i\beta))\). Multiplying gives \(\lambda^2-2\alpha\lambda+\alpha^2+\beta^2\). From the expansion of the characteristic polynomial, its coefficient of \(\lambda\) is \(-\operatorname{tr}\mathbf{A}\), and its constant term is \(\det\mathbf{A}\). Comparing coefficients yields \(\operatorname{tr}\mathbf{A}=2\alpha\) and \(\det\mathbf{A}=\alpha^2+\beta^2\). \(\square\)
Calculating a Nonreal Pair
Worked Example: A Nonreal Pair from Trace and Determinant
Find the eigenvalues of \(\mathbf{A}=\begin{bmatrix}1&-2\\2&1\end{bmatrix}\), and verify the result in the characteristic equation.
The trace is \(1+1=2\). The determinant is \(1\cdot1-(-2)\cdot2=1+4=5\). Therefore
Since \(\delta<0\), the eigenvalues are nonreal. Substituting into the formula gives \(\lambda=2/2\pm i\sqrt{16}/2=1\pm2i\). To check, the characteristic polynomial is \(\lambda^2-2\lambda+5\). For \(\lambda=1+2i\),
For \(\lambda=1-2i\), the calculation is \((-3-4i)-(2-4i)+5=0\). Both values satisfy the characteristic equation, as required.
When \(\delta<0\), writing the result as \(\alpha\pm i\beta\), with \(\beta>0\), makes the conjugate structure explicit. Here \(\alpha=\operatorname{tr}\mathbf{A}/2\) and \(\beta=\sqrt{-\delta}/2\). The next example applies the same calculation across an entire parameter range.
Worked Example: Classifying a Parameterized Family
For \(\mathbf{A}_t=\begin{bmatrix}t&-1\\1&0\end{bmatrix}\), where \(t\in\mathbb{R}\), determine for which values of \(t\) the eigenvalues are nonreal and calculate them in that case.
The trace is \(t+0=t\), and the determinant is \(t\cdot0-(-1)\cdot1=1\). Thus the discriminant is \(\delta=t^2-4\). It is negative exactly when \(-2<t<2\), because \(t^2<4\) is equivalent to \(|t|<2\). In this range, the eigenvalues are
For instance, at \(t=1\), this gives \(\lambda=\tfrac12\pm\tfrac{\sqrt{3}}{2}i\). The characteristic polynomial at \(t=1\) is \(\lambda^2-\lambda+1\). For \(\lambda=\tfrac12+\tfrac{\sqrt{3}}{2}i\), its square is \(-\tfrac12+\tfrac{\sqrt{3}}{2}i\), so \[ \lambda^2-\lambda+1 =\left(-\tfrac12+\tfrac{\sqrt{3}}{2}i\right) -\left(\tfrac12+\tfrac{\sqrt{3}}{2}i\right)+1=0. \] For the conjugate value, the imaginary parts have opposite signs and also cancel, giving zero.
At the endpoints \(t=\pm2\), \(\delta=0\), so the root is repeated and real. When \(|t|>2\), \(\delta>0\), so both roots are real and distinct. This confirms that the strict inequality \(-2<t<2\), not its endpoints, gives the nonreal pair.
Distinguishing the Three Discriminant Cases
Worked Example: A Repeated Real Root at the Boundary
For \(\mathbf{B}=\begin{bmatrix}2&1\\-1&0\end{bmatrix}\), determine the root type and verify the eigenvalue.
The trace is \(2+0=2\), and the determinant is \(2\cdot0-1\cdot(-1)=1\). Hence \(\delta=2^2-4(1)=0\). The repeated root is \(\lambda=2/2=1\). Directly,
Thus \(1\) is indeed the repeated characteristic root. The zero discriminant says that the root is repeated; it does not say whether there is one or more than one linearly independent eigenvector. That is a separate question.
Worked Example: Two Distinct Real Roots
Find and check the eigenvalues of \(\mathbf{C}=\begin{bmatrix}3&1\\1&1\end{bmatrix}\).
Its trace is \(3+1=4\), and its determinant is \(3\cdot1-1\cdot1=2\). Therefore \(\delta=4^2-4(2)=16-8=8>0\). The roots are \[ \lambda=\frac{4\pm\sqrt{8}}{2}=2\pm\sqrt{2}. \] The characteristic polynomial is \(\lambda^2-4\lambda+2\). Let \(\lambda=2+\sqrt{2}\). Then \(\lambda^2=6+4\sqrt{2}\), so \[ \lambda^2-4\lambda+2 =(6+4\sqrt{2})-(8+4\sqrt{2})+2=0. \] For \(\lambda=2-\sqrt{2}\), the square is \(6-4\sqrt{2}\), and substitution gives \((6-4\sqrt{2})-(8-4\sqrt{2})+2=0\). The roots are real and distinct, in agreement with the positive discriminant.
Common Mistake: Reversing the Sign Test
A frequent error is to treat a negative discriminant as evidence for real roots, or to reverse the order of the trace and determinant terms. For the first worked example, the correct calculation is \(\delta=(\operatorname{tr}\mathbf{A})^2-4\det\mathbf{A}=4-20=-16\), so the eigenvalues are nonreal. A mistaken calculation using \(4\det\mathbf{A}-(\operatorname{tr}\mathbf{A})^2\) gives \(16\) and might lead to the false conclusion that the roots are real.
The correction is to start from the characteristic polynomial and identify its discriminant in the usual order: for \(\lambda^2-T\lambda+D\), it is \((-T)^2-4D=T^2-4D\). A negative value gives the pair \(T/2\pm i\sqrt{-\delta}/2\). The determinant being positive does not settle the question: for example, the identity matrix has determinant \(1\), but its characteristic polynomial is \((\lambda-1)^2\), so its eigenvalues are real.
For a real \(2\times2\) matrix, the test can be summarized as follows:
| Discriminant | Characteristic roots | Conclusion |
|---|---|---|
| \(\delta>0\) | Two distinct real roots | No nonreal eigenvalues |
| \(\delta=0\) | One repeated real root | No nonreal eigenvalues |
| \(\delta<0\) | \(\tfrac12\operatorname{tr}\mathbf{A}\pm\tfrac12 i\sqrt{-\delta}\) | A nonreal conjugate pair |
This calculation is useful because trace and determinant summarize the quadratic characteristic equation without requiring a full eigenvector computation. The discriminant answers the root-type question; the formula then gives the pair exactly. It does not by itself determine eigenvectors or whether a matrix is diagonalizable.
Check Your Understanding
For each question, use the characteristic polynomial or the trace-determinant formula, and check any proposed roots by substitution.
- A real \(2\times2\) matrix has trace \(3\) and determinant \(4\). Calculate its discriminant and eigenvalues.
- For what real values of \(s\) is the discriminant of \(\begin{bmatrix}s&-2\\2&s\end{bmatrix}\) negative? Find the eigenvalues in that range.
- A real \(2\times2\) matrix has a nonreal eigenvalue with real part \(2\) and imaginary part \(3\). What are its trace and determinant?
- Explain why a positive determinant is not sufficient to conclude that a real \(2\times2\) matrix has nonreal eigenvalues.
- For a real \(2\times2\) matrix, what does a zero discriminant determine, and what does it not determine about eigenvectors?
