From a Matrix Family to a Root Classification
A parameter in a matrix can change its eigenvalues as the parameter varies. For a real \(2\times2\) matrix, the characteristic polynomial is quadratic, so its roots can be classified by one quantity: the discriminant. A positive discriminant gives two distinct real roots, a zero discriminant gives a repeated real root, and a negative discriminant gives no real roots. The main work is to calculate that quantity correctly and solve the resulting parameter condition.
Let \(\mathbf{A}=\begin{bmatrix}a&b\\c&d\end{bmatrix}\) be a real \(2\times2\) matrix. An eigenvalue is a scalar \(\lambda\) for which \(\mathbf{A}\mathbf{v}=\lambda\mathbf{v}\) for some nonzero vector \(\mathbf{v}\). The determinant criterion for eigenvalues says that this happens exactly when \(\det(\mathbf{A}-\lambda\mathbf{I}_2)=0\). Thus the eigenvalues are the roots of the characteristic polynomial \(p_{\mathbf{A}}(\lambda)=\det(\mathbf{A}-\lambda\mathbf{I}_2)\). Here we first classify its real roots; if the entries are real and the roots are not real, they are complex conjugates.
For a matrix family \(\mathbf{A}(t)\), the procedure is to compute \(p_{\mathbf{A}(t)}(\lambda)\), find its discriminant \(\Delta(t)\), and solve the inequalities or equation \(\Delta(t)>0\), \(\Delta(t)=0\), and \(\Delta(t)<0\). The parameter is real unless stated otherwise. A repeated root means that the characteristic polynomial has that root twice; it does not, by itself, tell us the dimension of the corresponding eigenspace.
The Discriminant in Terms of Trace and Determinant
Proof. Write \(\mathbf{A}=\begin{bmatrix}a&b\\c&d\end{bmatrix}\). Expanding the determinant gives \(\det(\mathbf{A}-\lambda\mathbf{I}_2)=(a-\lambda)(d-\lambda)-bc=\lambda^2-(a+d)\lambda+(ad-bc)\). Since \(\tau=a+d\) and \(\delta=ad-bc\), this is \(\lambda^2-\tau\lambda+\delta\). Its quadratic-formula discriminant is \((-\tau)^2-4\delta=\tau^2-4\delta\). Completing the square in the equation \(p_{\mathbf{A}}(\lambda)=0\) gives \((2\lambda-\tau)^2=\tau^2-4\delta=\Delta\). If \(\Delta>0\), there are exactly two choices \(2\lambda-\tau=\pm\sqrt{\Delta}\), giving two distinct real roots. If \(\Delta=0\), the only root is \(\lambda=\tau/2\), occurring twice. If \(\Delta<0\), the square of a real number cannot equal \(\Delta\), so there is no real root. This also gives the stated root formula when \(\Delta\geq0\). \(\square\)
This theorem makes the classification an algebraic problem in the parameter. In particular, the discriminant is not the determinant: it combines both trace and determinant. A determinant of zero tells us that zero is an eigenvalue; it does not, on its own, say whether the characteristic polynomial has distinct roots.
A Family With All Three Behaviours
Worked Example: Finding the Thresholds
For \(t\in\mathbb{R}\), classify the real roots of the characteristic polynomial of \(\mathbf{A}(t)=\begin{bmatrix}0&-1\\1&t\end{bmatrix}\).
Subtract \(\lambda\) from the diagonal and expand the determinant:
Its discriminant is \(\Delta(t)=(-t)^2-4(1)=t^2-4=(t-2)(t+2)\). The product is positive when \(t<-2\) or \(t>2\), zero when \(t=-2\) or \(t=2\), and negative when \(-2<t<2\). Therefore the polynomial has two distinct real roots outside the interval \([-2,2]\), a repeated real root at its endpoints, and no real roots inside the interval.
For a check in the distinct-root case, take \(t=3\). The formula gives \(\lambda=(3\pm\sqrt{5})/2\). If \(r=(3\pm\sqrt{5})/2\), then \(r^2=(7\pm3\sqrt{5})/2\) and \(3r=(9\pm3\sqrt{5})/2\). Hence \(r^2-3r+1=0\), as required. At \(t=2\), the polynomial is \(\lambda^2-2\lambda+1\); substituting its repeated root \(\lambda=1\) gives \(1-2+1=0\). For \(t=0\), it is \(\lambda^2+1\), which is positive for every real \(\lambda\), so it has no real root. These checks agree with the discriminant classification.
The factorization of \(\Delta(t)\) made the threshold values visible. When a discriminant is not already factored, it may be useful to expand it and then complete the square or factor it before describing the parameter ranges. The conclusion must come from the sign over the whole range of parameters, not just from a few sample values.
Families That Stay in One Root Regime
Worked Example: Distinct Real Roots for Every Parameter
Classify the real roots for \(\mathbf{B}(t)=\begin{bmatrix}t&1\\1&-t\end{bmatrix}\), where \(t\in\mathbb{R}\).
The trace is \(t+(-t)=0\), and the determinant is \(t(-t)-1\cdot1=-t^2-1\). Thus
The discriminant is \(\Delta(t)=0^2-4(-t^2-1)=4(t^2+1)\). Since \(t^2\geq0\), we have \(t^2+1>0\) for every real \(t\), so the roots are always distinct and real. The formula gives \(\lambda=\pm\sqrt{t^2+1}\). To check, for either choice \(r\), \(r^2=t^2+1\), and therefore \(p_{\mathbf{B}(t)}(r)=r^2-t^2-1=0\).
Worked Example: No Real Roots for Any Parameter
Classify the real roots for \(\mathbf{C}(t)=\begin{bmatrix}t&-2\\2&t\end{bmatrix}\), where \(t\in\mathbb{R}\).
The trace is \(2t\), and the determinant is \(t^2-(-2)(2)=t^2+4\). Consequently,
The discriminant is \((2t)^2-4(t^2+4)=-16\), which is negative for every real \(t\). Thus there are no real roots for any parameter value. Over \(\mathbb{C}\), the roots are \(t+2i\) and \(t-2i\). Substituting either root into \((\lambda-t)^2+4\) gives \((\pm2i)^2+4=-4+4=0\), verifying both roots.
Why the Sign Test Works for a Parameter
The classification is an if-and-only-if rule at each fixed parameter value, not a guess based on a graph or a few substitutions. The theorem supplies both directions: a positive discriminant gives two distinct real roots, and two distinct real roots require a positive discriminant. The same equivalence holds for a repeated root and a zero discriminant. Thus the parameter analysis can be organized as a short sequence of justified steps.
Evaluate \(\det(\mathbf{A}(t)-\lambda\mathbf{I}_2)\), keeping \(\lambda\) as the polynomial variable and \(t\) as the parameter.
For \(\lambda^2+B(t)\lambda+C(t)\), calculate \(\Delta(t)=B(t)^2-4C(t)\). Equivalently, use \(\operatorname{tr}(\mathbf{A}(t))^2-4\det(\mathbf{A}(t))\).
Solve the strict inequalities and equality separately. The parameter values with \(\Delta>0\), \(\Delta=0\), and \(\Delta<0\) give the three respective root cases.
When roots are claimed, substitute them into the characteristic polynomial. When no real roots are claimed, justify that the discriminant is negative throughout the stated range.
In the first family, for instance, the sign analysis was not just that \(t^2-4\) is negative at \(t=0\). It was the factorization \((t-2)(t+2)\) that established exactly where the sign changes. For a more complicated expression, completing the square can establish a sign that is not obvious from its expanded form.
Common Mistake: Using the Determinant as the Discriminant
A tempting but incorrect shortcut is to inspect \(\det\mathbf{A}\) and treat its sign as the test for distinct, repeated, or nonreal roots. For example, in the first family at \(t=0\), the matrix is \(\begin{bmatrix}0&-1\\1&0\end{bmatrix}\) and its determinant is \(1\). The incorrect calculation “the determinant is positive, so there are two distinct real eigenvalues” confuses two different quantities. The characteristic polynomial is \(\lambda^2+1\), and its discriminant is \(0^2-4(1)=-4\), so it has no real roots.
The determinant is the constant term of a \(2\times2\) characteristic polynomial, while the discriminant also depends on the coefficient of \(\lambda\). Use \(\Delta=(\operatorname{tr}\mathbf{A})^2-4\det\mathbf{A}\) for root classification. If the question is whether zero is an eigenvalue, then the determinant is relevant: \(p_{\mathbf{A}}(0)=\det\mathbf{A}\), so zero is a root exactly when the determinant is zero.
Finally, a repeated characteristic root identifies a repeated root of the polynomial, not the number of independent eigenvectors. Determining eigenspace dimensions requires a separate calculation and is not part of the discriminant test.
Check Your Understanding
For each family, compute the characteristic polynomial before classifying its real roots.
- For \(\mathbf{D}(t)=\begin{bmatrix}t&1\\-1&0\end{bmatrix}\), find the parameter values giving distinct real roots, a repeated root, and no real roots.
- For \(\mathbf{E}(t)=\begin{bmatrix}t&0\\0&-t\end{bmatrix}\), classify the roots for \(t=0\) and for \(t\ne0\).
- Explain why a negative discriminant rules out every real eigenvalue, rather than merely ruling out two distinct real eigenvalues.
- For a real \(2\times2\) matrix with trace \(5\) and determinant \(6\), calculate the discriminant and classify the roots.
- Give a specific reason that \(\det\mathbf{A}>0\) alone cannot establish that a real \(2\times2\) matrix has two real eigenvalues.
