The Discriminant And Real Eigenvalues Of Two-By-Two Matrices

What the Discriminant Tells Us

A real two-by-two matrix can have two distinct real eigenvalues, one real eigenvalue that is repeated, or no real eigenvalues. The characteristic polynomial records the eigenvalues as its roots, so the quadratic discriminant provides a direct way to distinguish these possibilities. This is a classification of the eigenvalues themselves: it does not, by itself, tell us the dimensions of their eigenspaces.

Throughout, let \(\mathbf{A}\in\mathbb{R}^{2\times2}\). A real number \(\lambda\) is an eigenvalue of \(\mathbf{A}\) if there is a nonzero vector \(\mathbf{v}\in\mathbb{R}^2\) such that \(\mathbf{A}\mathbf{v}=\lambda\mathbf{v}\). Equivalently, \(\mathbf{A}-\lambda\mathbf{I}_2\) is singular, so its determinant is zero. We use the course convention \(p_{\mathbf{A}}(\lambda)=\det(\mathbf{A}-\lambda\mathbf{I}_2)\); in particular, the eigenvalues are the roots of this polynomial.

To apply this definition, first compute the characteristic polynomial using the stated sign convention. Expanding the two-by-two determinant gives

Thus the coefficient of \(\lambda\) is \(-(a+d)\), and the constant term is \(ad-bc\). Substitution into the quadratic discriminant formula gives \(\Delta=(a+d)^2-4(ad-bc)\). Each term in this calculation matters: in particular, the determinant expansion subtracts \(bc\).

The Entrywise Formula and Its Proof

Proof. From the characteristic polynomial just computed, the discriminant is \((a+d)^2-4(ad-bc)\). Expanding and collecting terms gives

This proves the identity for every choice of real entries \(a,b,c,d\). \(\square\)

This entrywise form is useful because it makes the possible contributions to \(\Delta\) visible. The square \((a-d)^2\) is nonnegative, but \(4bc\) can be positive, zero, or negative. Therefore, the sign cannot be determined from the diagonal entries alone in every case: the product of the off-diagonal entries can change the result.

Classifying the Real Eigenvalues

Proof. Write \(p_{\mathbf{A}}(\lambda)=\lambda^2-(a+d)\lambda+(ad-bc)\). The quadratic formula gives its roots as

If \(\Delta>0\), then \(\sqrt{\Delta}\) is a positive real number. The two choices of sign produce real numbers, and they are distinct because their difference is \(\sqrt{\Delta}\ne0\). Each root makes \(p_{\mathbf{A}}(\lambda)=\det(\mathbf{A}-\lambda\mathbf{I}_2)=0\), so each is a real eigenvalue.

If \(\Delta=0\), both choices give the same real number \(\lambda=(a+d)/2\). Completing the square, or substituting the zero discriminant into the polynomial, gives \(p_{\mathbf{A}}(\lambda)=(\lambda-(a+d)/2)^2\). Hence this real root has algebraic multiplicity two, and there is only one distinct real eigenvalue.

If \(\Delta<0\), the square root in the quadratic formula is not real. More explicitly, for real \(\lambda\), completing the square gives

because the first term is nonnegative and \(-\Delta/4>0\). Thus the polynomial has no real root. Since real eigenvalues are exactly its real roots, \(\mathbf{A}\) has no real eigenvalues. These three cases exhaust the possibilities for the real number \(\Delta\). \(\square\)

The cases describe the roots counted as polynomial roots, not the number of linearly independent eigenvectors. In particular, a repeated real root has algebraic multiplicity two, but the discriminant calculation alone does not determine the dimension of its eigenspace. That requires examining the corresponding equation \((\mathbf{A}-\lambda\mathbf{I}_2)\mathbf{v}=\mathbf{0}\).

Worked Classifications

Reading the Sign from the Entries

The identity \(\Delta=(a-d)^2+4bc\) also gives a useful way to assess the cases before computing the full characteristic polynomial. If \(bc>0\), then \(4bc>0\), so \(\Delta>0\) and there are two distinct real eigenvalues. If \(bc=0\), then \(\Delta=(a-d)^2\): it is positive when \(a\ne d\), and zero when \(a=d\). If \(bc<0\), the positive square term competes with a negative term, so either sign or zero is possible.

For \(bc<0\), write \(4bc=-4|bc|\). Then \(\Delta=(a-d)^2-4|bc|\). It is positive when \((a-d)^2>4|bc|\), zero when \((a-d)^2=4|bc|\), and negative when \((a-d)^2<4|bc|\). This comparison is another way to apply the same discriminant test, not a separate test that bypasses the characteristic polynomial.

For example, in \(\mathbf{B}\), \(bc=-1\) and \((a-d)^2=4\), so the two contributions exactly cancel and \(\Delta=0\). In \(\mathbf{C}\), \(bc=-1\) but \((a-d)^2=0\), so the negative contribution remains and \(\Delta<0\). These checks show why a negative product \(bc\) does not automatically mean there are no real eigenvalues: the size of the diagonal difference matters as well.

Common Mistake: Losing the Determinant Sign

A frequent error is to use \(+bc\) instead of \(-bc\) when expanding \(\det\begin{bmatrix}x&b\\c&y\end{bmatrix}=xy-bc\). For the matrix \(\mathbf{C}\), that mistake would give the false polynomial \(\lambda^2-1\), leading to the incorrect conclusion that its eigenvalues are \(1\) and \(-1\).

The correction is to expand the determinant using the product of the main diagonal entries minus the product of the other diagonal entries:

The corrected polynomial is positive for every real \(\lambda\), in agreement with \(\Delta=-4\). A reliable check is to substitute the proposed value into \(\det(\mathbf{A}-\lambda\mathbf{I}_2)\). The false candidates \(1\) and \(-1\) both give determinant \(2\), not zero, so neither is an eigenvalue.

What This Test Does—and Does Not—Answer

The discriminant test is useful when the immediate question is whether the eigenvalues of a real two-by-two matrix are real and whether they are distinct. It requires only the entries of the matrix and exact arithmetic. Once \(\Delta\) is known, the classification is complete:

The classification should not be stretched beyond what it proves. A zero discriminant says that the characteristic polynomial has a repeated real root; it does not establish how many independent eigenvectors correspond to that root. Nor does the sign alone provide the eigenvectors. Those questions require solving the eigenvalue equation for the relevant value of \(\lambda\). Keeping the root classification separate from the eigenspace calculation prevents a common logical leap from “repeated eigenvalue” to an unsupported claim about eigenvectors.