Perron-Frobenius for Positive Two-by-Two Matrices

Why Positivity Controls the Eigenvalues

For a general matrix, eigenvalues can be real or complex, and the entries alone may reveal little about their sizes or eigenvectors. A strictly positive two-by-two matrix is a special case where the characteristic polynomial makes the full picture accessible. Its entries force two real, distinct eigenvalues; one is positive and strictly larger in magnitude than the other. The eigenvector for this dominant eigenvalue can be chosen with both entries positive.

Here, “strictly positive” refers to every matrix entry, not merely to a nonnegative entry. We work over \(\mathbb{R}\). Recall that an eigenvalue \(\lambda\) of a matrix \(\mathbf{A}\) is a scalar for which \(\mathbf{A}\mathbf{x}=\lambda\mathbf{x}\) has a solution \(\mathbf{x}\ne\mathbf{0}\); such a nonzero \(\mathbf{x}\) is an eigenvector. A left eigenvector will be written as a column \(\mathbf{y}\) satisfying \(\mathbf{A}^{\top}\mathbf{y}=\lambda\mathbf{y}\), so that \(\mathbf{y}^{\top}\mathbf{A}=\lambda\mathbf{y}^{\top}\).

Finding the Two Characteristic Roots

Write a strictly positive matrix as \(\mathbf{A}=\begin{bmatrix}a&b\\c&d\end{bmatrix}\), where \(a,b,c,d>0\). Its characteristic polynomial, using the convention \(p_{\mathbf{A}}(\lambda)=\det(\mathbf{A}-\lambda\mathbf{I}_2)\), is

The quadratic formula gives two roots. It is useful to label the larger one \(\lambda_+\) and the smaller one \(\lambda_-\):

Because \(b,c>0\), the quantity under the square root is strictly positive. Thus the roots are real and distinct. Moreover, \(\sqrt{(a-d)^2+4bc}>|a-d|\). This gives \(\lambda_+>\max(a,d)>0\) and \(\lambda_-<\min(a,d)\). The second inequality does not imply that \(\lambda_-\) is negative: depending on the entries, it can be positive, zero, or negative.

Proof. The characteristic polynomial computation and quadratic formula give the displayed roots. Since \(bc>0\), their discriminant \((a-d)^2+4bc\) is positive, so they are distinct. In particular, each is a simple root of the characteristic polynomial, and \(\lambda_+\) has algebraic multiplicity one. Also \(\sqrt{(a-d)^2+4bc}>|a-d|\), so \(\lambda_+>a>0\) and \(\lambda_+>d>0\). The difference and sum of the roots are \(\lambda_+-\lambda_-=\sqrt{(a-d)^2+4bc}>0\) and \(\lambda_++\lambda_-=a+d>0\). If \(\lambda_-\ge0\), the difference gives \(\lambda_+>|\lambda_-|\). If \(\lambda_-<0\), the sum gives \(\lambda_+-|\lambda_-|=\lambda_++\lambda_->0\). Thus the magnitude inequality holds in both cases.

Set \(\mathbf{v}=\begin{bmatrix}b\\\lambda_+-a\end{bmatrix}\). Both entries are positive, since \(b>0\) and \(\lambda_+>a\). The first entry of \(\mathbf{A}\mathbf{v}\) is \(ab+b(\lambda_+-a)=b\lambda_+\). Since \(\lambda_+\) is a root of the characteristic polynomial, \((\lambda_+-a)(\lambda_+-d)=bc\). Therefore the second entry is

Consequently \(\mathbf{A}\mathbf{v}=\lambda_+\mathbf{v}\). Applying the same construction to \(\mathbf{A}^{\top}=\begin{bmatrix}a&c\\b&d\end{bmatrix}\), which also has strictly positive entries and the same characteristic polynomial, gives a strictly positive \(\mathbf{y}\) with \(\mathbf{A}^{\top}\mathbf{y}=\lambda_+\mathbf{y}\). Finally, if \(\mathbf{x}=\begin{bmatrix}x_1\\x_2\end{bmatrix}\) has both entries positive and is an eigenvector, its eigenvalue cannot be \(\lambda_-\): the first row equation would give \((a-\lambda_-)x_1+bx_2=0\), although both terms on the left are positive because \(a-\lambda_->0\). Hence its eigenvalue is \(\lambda_+\). The first row equation for that eigenvalue gives \(x_2=(\lambda_+-a)x_1/b\), so every such \(\mathbf{x}\) is a positive scalar multiple of \(\mathbf{v}\). \(\square\)

The magnitude comparison is the central point: “dominant” here means strictly largest in absolute value, not merely larger as a real number. The proof also covers the case where both eigenvalues are positive; in that case the difference \(\lambda_+-\lambda_->0\) is the comparison needed. When the second eigenvalue is negative, its sign must instead be handled using the positive sum \(\lambda_++\lambda_-\).

Worked Examples

Common Pitfall: Comparing the Roots Incorrectly

A frequent error is to infer that the second eigenvalue must be positive because all four entries are positive, or to compare the eigenvalues only by their ordinary order. The second worked example directly corrects both errors: for \(\mathbf{A}=\begin{bmatrix}1&4\\4&1\end{bmatrix}\), the calculation gives \(\lambda_+=5\) and \(\lambda_-=-3\). The statement \(5>-3\) is true but is not by itself the dominance claim; the required check is \(5>|-3|=3\). The proof establishes that absolute-value inequality in either sign case.

Strict positivity is also an essential hypothesis. For example, \(\begin{bmatrix}2&0\\0&1\end{bmatrix}\) has nonnegative entries but is not strictly positive. Its eigenvector for \(2\) is a multiple of \(\begin{bmatrix}1\\0\end{bmatrix}\), not a vector with both entries positive. Thus the conclusion about a strictly positive eigenvector cannot simply be transferred to every matrix with nonnegative entries.

What the Two-by-Two Result Gives

The result is a complete description of the eigenvalue magnitudes in this dimension: there are exactly two distinct real eigenvalues, \(\lambda_+\) and \(\lambda_-\), and \(\lambda_+\) is strictly dominant. Its positive right eigenvector is unique up to multiplication by a positive scalar among vectors with positive entries. Applying the same reasoning to the transpose supplies a positive left eigenvector for that same eigenvalue. These statements use strict positivity of all four entries; changing even one entry to zero means the theorem’s hypotheses no longer apply.

The explicit formulas make the proof particularly transparent. Positivity of \(b\) and \(c\) makes the discriminant exceed \((a-d)^2\), which places \(\lambda_+\) above both diagonal entries and \(\lambda_-\) below both. That placement gives the positive eigenvector and rules out a positive eigenvector for the smaller root. The sum and difference of the roots then finish the magnitude comparison without assuming in advance whether the smaller root is positive or negative.