Eigenvalues of Symmetric 2x2 Matrices

Set Up the 2×2 Symmetric Matrix

A real symmetric 2×2 matrix has equal off-diagonal entries, so it can be written as \(\mathbf{A}=\begin{bmatrix}a&b\\b&d\end{bmatrix}\), where \(a,b,d\in\mathbb{R}\). Its entries determine both the characteristic polynomial and the equations that define its eigenspaces. In dimension two, these calculations can be made explicit, including the special case in which both eigenvalues coincide.

A real number \(\lambda\) is an eigenvalue of \(\mathbf{A}\) if there is a nonzero vector \(\mathbf{v}\in\mathbb{R}^2\) such that \(\mathbf{A}\mathbf{v}=\lambda\mathbf{v}\). The eigenspace for \(\lambda\), written \(E_{\lambda}\), consists of the zero vector together with all eigenvectors for \(\lambda\); equivalently, \(E_{\lambda}=\operatorname{Nul}(\mathbf{A}-\lambda\mathbf{I}_2)\). The characteristic polynomial, with the course convention for its sign, is \(p_{\mathbf{A}}(\lambda)=\det(\mathbf{A}-\lambda\mathbf{I}_2)\).

To see the polynomial, subtract \(\lambda\) from both diagonal entries and take the 2×2 determinant:

The roots of this polynomial are precisely the eigenvalues. In this 2×2 setting, the reason is that \(\lambda\) is an eigenvalue exactly when \(\mathbf{A}-\lambda\mathbf{I}_2\) has a nonzero vector in its null space, which is equivalent to its determinant being zero. For a 2×2 matrix, determinant zero means its columns are linearly dependent, so they have a nontrivial linear combination equal to \(\mathbf{0}\); conversely, a nonzero null vector gives such a dependence and determinant zero.

Compute the Eigenvalues

Applying the quadratic formula to \(p_{\mathbf{A}}(\lambda)=0\) gives two roots. Rewriting the discriminant makes the role of symmetry especially visible:

Because \((a-d)^2+4b^2\geq 0\), both roots are real. If this quantity is positive, there are two distinct eigenvalues. It is zero precisely when \(a=d\) and \(b=0\): each of the nonnegative terms \((a-d)^2\) and \(4b^2\) must be zero. In that case \(\mathbf{A}=a\mathbf{I}_2\), and the characteristic polynomial has a repeated root. This gives a useful distinction: a repeated root does not by itself mean there is only one independent direction of eigenvectors.

Proof. The determinant calculation gives \(p_{\mathbf{A}}(\lambda)=\lambda^2-(a+d)\lambda+(ad-b^2)\). The quadratic formula therefore gives the displayed roots. Their discriminant is \((a-d)^2+4b^2\), which is nonnegative, so both roots are real. The roots coincide exactly when the discriminant is zero. Since both summands are nonnegative real numbers, their sum is zero exactly when \(a-d=0\) and \(b=0\). Then \(\mathbf{A}=a\mathbf{I}_2\); for every \(\mathbf{v}\in\mathbb{R}^2\), \(\mathbf{A}\mathbf{v}=a\mathbf{v}\). Thus every vector is in \(E_a\), including \(\mathbf{0}\), and \(E_a=\mathbb{R}^2\). \(\square\)

Find the Eigenspaces Without Losing a Case

For a nonzero off-diagonal entry \(b\), an eigenvector can be written directly from the first row of \((\mathbf{A}-\lambda\mathbf{I}_2)\mathbf{v}=\mathbf{0}\). This formula works for either eigenvalue. The nonzero condition on \(b\) matters: if \(b=0\), the formula would produce the zero vector for one of the eigenvalues of a diagonal matrix, so that case must be handled separately.

Proof. First suppose \(b\ne0\), and let \(\lambda\) be an eigenvalue. The vector \(\mathbf{v}=(b,\lambda-a)^{\top}\) is nonzero because its first coordinate is nonzero. Its first coordinate after multiplication by \(\mathbf{A}-\lambda\mathbf{I}_2\) is \((a-\lambda)b+b(\lambda-a)=0\). Since \(\lambda\) is an eigenvalue, \((a-\lambda)(d-\lambda)-b^2=0\), so \(b^2=(a-\lambda)(d-\lambda)\). The second coordinate is \(b^2+(d-\lambda)(\lambda-a)=b^2-(d-\lambda)(a-\lambda)=0\). Hence \(\mathbf{v}\in E_\lambda\). Conversely, any \((x,y)^{\top}\in E_\lambda\) satisfies \((a-\lambda)x+by=0\). Since \(b\ne0\), this gives \(y=(\lambda-a)x/b\), so \((x,y)^{\top}=(x/b)(b,\lambda-a)^{\top}\). This proves the stated span. Now suppose \(b=0\), so \(\mathbf{A}\) is diagonal. If \(a\ne d\), then \(\mathbf{A}(x,y)^{\top}=(ax,dy)^{\top}\); the equation for eigenvalue \(a\) forces \(y=0\), and the equation for \(d\) forces \(x=0\). The corresponding eigenspaces are the two stated spans. If \(a=d\), the matrix is \(a\mathbf{I}_2\), and every vector satisfies \(\mathbf{A}\mathbf{v}=a\mathbf{v}\). \(\square\)

Common Errors and a Reliable Procedure

A frequent error is to omit the \(-b^2\) term when expanding the determinant. For \(\mathbf{A}=\begin{bmatrix}1&2\\2&1\end{bmatrix}\), an incorrect calculation might use only the product of the diagonal entries and write \(p_{\mathbf{A}}(\lambda)=(1-\lambda)^2\), leading to the false eigenvalue \(1\). The off-diagonal product must be subtracted: the correct polynomial is \((1-\lambda)^2-2^2=\lambda^2-2\lambda-3\), whose roots are \(3\) and \(-1\). Indeed, \(\mathbf{A}-\mathbf{I}_2=\begin{bmatrix}0&2\\2&0\end{bmatrix}\) has only the zero vector in its null space, so \(1\) is not an eigenvalue.

A second error is to infer that a repeated eigenvalue has a one-dimensional eigenspace. The repeated-root condition for a real symmetric 2×2 matrix forces the matrix to be a scalar multiple of the identity. Every vector is then an eigenvector, and the eigenspace is all of \(\mathbb{R}^2\). By contrast, when \(b\ne0\), the proposition shows that each eigenspace is spanned by one nonzero vector.

For a reliable calculation, first expand \(\det(\mathbf{A}-\lambda\mathbf{I}_2)\), keeping the off-diagonal product. Next solve the resulting quadratic and check whether the discriminant is zero. Finally, solve \((\mathbf{A}-\lambda\mathbf{I}_2)\mathbf{v}=\mathbf{0}\) for each eigenvalue, using the nonzero-\(b\) formula only when \(b\ne0\), and multiply \(\mathbf{A}\mathbf{v}\) to verify each proposed generator.

Why the Cases Matter

These calculations distinguish three situations that can look similar if one considers only the polynomial: two distinct eigenvalues with nonzero off-diagonal entries, a diagonal matrix with distinct diagonal entries, and a scalar matrix with a repeated eigenvalue. The eigenvalue formula applies to all three, but the safest eigenspace calculation depends on which situation occurs. Checking the entries before dividing or selecting a proposed eigenvector prevents undefined steps and zero-vector answers.