Characteristic Polynomial Sign Conventions

Two Conventions for the Same Eigenvalue Test

There are two common ways to write the determinant used to find eigenvalues. One places the scalar parameter first, as \(\det(\lambda\mathbf{I}_n-\mathbf{A})\); the other places the matrix first, as \(\det(\mathbf{A}-\lambda\mathbf{I}_n)\). Both are legitimate, but their polynomial values can differ by a sign. That sign depends on the dimension, so a formula copied from a source using the other convention may look different even when it describes exactly the same eigenvalues.

Let \(\mathbf{A}\in\mathbb{F}^{n\times n}\), where \(\mathbb{F}=\mathbb{R}\) or \(\mathbb{C}\) and \(n\geq 1\). Write \(\mathbf{I}_n\) for the \(n\times n\) identity matrix. In the convention used here, the characteristic polynomial is \(p_{\mathbf{A}}(\lambda)=\det(\mathbf{A}-\lambda\mathbf{I}_n)\). To compare the alternative convention, define \(q_{\mathbf{A}}(\lambda)=\det(\lambda\mathbf{I}_n-\mathbf{A})\). We consider roots in \(\mathbb{C}\); if the entries and parameter are restricted to \(\mathbb{R}\), the same determinant identity holds over \(\mathbb{R}\).

Definition: For a square matrix \(\mathbf{A}\in\mathbb{F}^{n\times n}\), the two characteristic-polynomial conventions are \(p_{\mathbf{A}}(\lambda)=\det(\mathbf{A}-\lambda\mathbf{I}_n)\) and \(q_{\mathbf{A}}(\lambda)=\det(\lambda\mathbf{I}_n-\mathbf{A})\). A scalar \(\lambda\) is a root of either polynomial when that polynomial evaluates to zero.

These conventions are related to eigenvalues through the eigenvalue equation: \(\lambda\) is an eigenvalue of \(\mathbf{A}\) precisely when \(\mathbf{A}-\lambda\mathbf{I}_n\) is singular, or equivalently when its determinant is zero. Thus the roots of \(p_{\mathbf{A}}\) are the eigenvalues. The determinant identity below will show that \(q_{\mathbf{A}}\) has the same roots, even though its values may have the opposite sign.

The Dimension-Dependent Sign

Theorem (Sign relation between conventions): For every \(\mathbf{A}\in\mathbb{F}^{n\times n}\) and every \(\lambda\in\mathbb{F}\), \(q_{\mathbf{A}}(\lambda)=(-1)^n p_{\mathbf{A}}(\lambda)\). Consequently, the conventions agree when \(n\) is even and are negatives of each other when \(n\) is odd.

Proof. The matrices in the two determinants satisfy \(\lambda\mathbf{I}_n-\mathbf{A}=-(\mathbf{A}-\lambda\mathbf{I}_n)\). Multiplying a matrix by \(-1\) multiplies each of its \(n\) rows by \(-1\). Since scaling one row by a scalar \(c\) scales the determinant by \(c\), scaling all \(n\) rows by \(-1\) scales it by \((-1)^n\). Therefore \(\det(\lambda\mathbf{I}_n-\mathbf{A})=(-1)^n\det(\mathbf{A}-\lambda\mathbf{I}_n)\), which is the claimed relation. If \(n\) is even then \((-1)^n=1\); if \(n\) is odd then \((-1)^n=-1\). \(\square\)

The sign is determined by the number of rows, not by the entries of \(\mathbf{A}\) or by the value of \(\lambda\). For instance, changing the entries changes both polynomials, but it does not change whether the two conventions agree or differ by a minus sign. The distinction is between even and odd dimensions: it is not enough to remember that the matrices inside the determinants are negatives of each other without also counting how many rows are scaled.

Worked Example: A One-Dimensional Matrix

Let \(\mathbf{A}=\begin{bmatrix}3\end{bmatrix}\). Compute both conventions and compare their roots.

Here \(n=1\), so the sign relation predicts that the polynomials are negatives. Directly,

p_{\mathbf{A}}(\lambda)=\det\begin{bmatrix}3-\lambda\end{bmatrix}=3-\lambda, \qquad q_{\mathbf{A}}(\lambda)=\det\begin{bmatrix}\lambda-3\end{bmatrix}=\lambda-3.

Indeed, \(q_{\mathbf{A}}(\lambda)=-(3-\lambda)=-p_{\mathbf{A}}(\lambda)\). Both polynomials have the root \(\lambda=3\): \(p_{\mathbf{A}}(3)=3-3=0\), and \(q_{\mathbf{A}}(3)=3-3=0\).

Check: substituting \(\lambda=3\) into the matrix \(\mathbf{A}-\lambda\mathbf{I}_1\) gives \(\begin{bmatrix}0\end{bmatrix}\), whose determinant is \(0\), as required for the eigenvalue test. The two polynomial values differ by a sign at every parameter, but their zero occurs at the same parameter.

Worked Example: A Two-Dimensional Matrix

Let \(\mathbf{A}=\begin{bmatrix}1&2\\3&4\end{bmatrix}\). Compute both polynomials and check the predicted sign and their roots.

For the convention \(p_{\mathbf{A}}(\lambda)=\det(\mathbf{A}-\lambda\mathbf{I}_2)\),

\begin{aligned} p_{\mathbf{A}}(\lambda) &=\det\begin{bmatrix}1-\lambda&2\\3&4-\lambda\end{bmatrix}\\ &=(1-\lambda)(4-\lambda)-2\cdot 3\\ &=4-5\lambda+\lambda^2-6\\ &=\lambda^2-5\lambda-2. \end{aligned}

For the other convention,

\begin{aligned} q_{\mathbf{A}}(\lambda) &=\det\begin{bmatrix}\lambda-1&-2\\-3&\lambda-4\end{bmatrix}\\ &=(\lambda-1)(\lambda-4)-(-2)(-3)\\ &=\lambda^2-5\lambda+4-6\\ &=\lambda^2-5\lambda-2. \end{aligned}

Since \(n=2\) is even, the theorem predicts equality, and the calculations confirm \(q_{\mathbf{A}}=p_{\mathbf{A}}\). Solving \(\lambda^2-5\lambda-2=0\) gives \(\lambda=(5\pm\sqrt{33})/2\). Substitution verifies the roots: for either choice, \(\lambda^2=5\lambda+2\), so \(\lambda^2-5\lambda-2=0\). Each root therefore makes both determinant conventions zero.

Check: at \(\lambda=0\), both formulas give \(-2\). Directly, \(p_{\mathbf{A}}(0)=\det\mathbf{A}=1\cdot4-2\cdot3=-2\), while \(q_{\mathbf{A}}(0)=\det(-\mathbf{A})=(-1)^2\det\mathbf{A}=-2\).

The Roots Do Not Change

Corollary (Same roots): For any square matrix \(\mathbf{A}\in\mathbb{F}^{n\times n}\), the two characteristic-polynomial conventions have exactly the same roots.

Proof. Fix a scalar \(\lambda\). By the sign relation, \(q_{\mathbf{A}}(\lambda)=(-1)^n p_{\mathbf{A}}(\lambda)\). The scalar \((-1)^n\) is either \(1\) or \(-1\), and is never zero. If \(p_{\mathbf{A}}(\lambda)=0\), then \(q_{\mathbf{A}}(\lambda)=(-1)^n\cdot 0=0\). Conversely, if \(q_{\mathbf{A}}(\lambda)=0\), then \(p_{\mathbf{A}}(\lambda)=(-1)^n q_{\mathbf{A}}(\lambda)=0\), because \(((-1)^n)^2=1\). Thus each root of either polynomial is a root of the other. \(\square\)

This result means the conventions do not require different eigenvalue calculations or a change of parameter. The same value of \(\lambda\) is tested in both polynomials. The relation multiplies the polynomial by a nonzero constant; it does not replace \(\lambda\) by \(-\lambda\). This distinction matters when translating a factored polynomial or a list of roots from one convention to the other.

Worked Example: A Three-Dimensional Matrix

Let \(\mathbf{A}=\begin{bmatrix}0&1&0\\0&0&1\\6&-11&6\end{bmatrix}\). Compute the characteristic polynomial in each convention and identify the roots.

First, expand \(\det(\mathbf{A}-\lambda\mathbf{I}_3)\) along the first row:

\begin{aligned} p_{\mathbf{A}}(\lambda) &=\det\begin{bmatrix}-\lambda&1&0\\0&-\lambda&1\\6&-11&6-\lambda\end{bmatrix}\\ &=(-\lambda)\bigl((-\lambda)(6-\lambda)+11\bigr)+6\\ &=-\lambda(\lambda^2-6\lambda+11)+6\\ &=-\lambda^3+6\lambda^2-11\lambda+6. \end{aligned}

For the other convention, expansion along the first row gives

\begin{aligned} q_{\mathbf{A}}(\lambda) &=\det\begin{bmatrix}\lambda&-1&0\\0&\lambda&-1\\-6&11&\lambda-6\end{bmatrix}\\ &=\lambda\bigl(\lambda(\lambda-6)+11\bigr)-6\\ &=\lambda^3-6\lambda^2+11\lambda-6\\ &=(\lambda-1)(\lambda-2)(\lambda-3). \end{aligned}

In dimension \(3\), the sign relation predicts \(p_{\mathbf{A}}=-q_{\mathbf{A}}\). Indeed, \(p_{\mathbf{A}}(\lambda)=-(\lambda-1)(\lambda-2)(\lambda-3)\). The roots are \(1,2,3\) under either convention: each makes one factor in \(q_{\mathbf{A}}\) zero and therefore also makes \(p_{\mathbf{A}}\) zero.

Check: at \(\lambda=1\), the second row of \(\mathbf{A}-\lambda\mathbf{I}_3\) is \((0,-1,1)\), and its determinant evaluates to \(-1+6-11+6=0\) by the polynomial formula. At the same value, \(q_{\mathbf{A}}(1)=(1-1)(1-2)(1-3)=0\). The polynomials have opposite values away from their roots, but both vanish at \(\lambda=1\).

Why One Convention Is Often Monic

The sign relation also explains a common presentation choice. The convention \(q_{\mathbf{A}}(\lambda)=\det(\lambda\mathbf{I}_n-\mathbf{A})\) is monic: its highest-degree term is \(\lambda^n\). The convention \(p_{\mathbf{A}}(\lambda)=\det(\mathbf{A}-\lambda\mathbf{I}_n)\) has highest-degree term \((-1)^n\lambda^n\). Here “monic” means that the coefficient of the highest power is \(1\).

To see the leading term, write the \(j\)th column of \(\lambda\mathbf{I}_n-\mathbf{A}\) as \(\lambda\mathbf{e}_j-\mathbf{a}_j\), where \(\mathbf{e}_j\) is the \(j\)th standard basis vector and \(\mathbf{a}_j\) is the \(j\)th column of \(\mathbf{A}\). By multilinearity of the determinant in its columns, the term of degree \(n\) comes from selecting \(\lambda\mathbf{e}_j\) in every column. Its value is \(\lambda^n\det(\mathbf{I}_n)=\lambda^n\). Every other term uses at least one column \(-\mathbf{a}_j\), so it has degree at most \(n-1\). Hence \(q_{\mathbf{A}}\) is monic. The sign relation then gives the leading coefficient of \(p_{\mathbf{A}}\) as \((-1)^n\).

This is a useful check on a computation. If a calculation using \(q_{\mathbf{A}}\) produces a polynomial whose leading coefficient is not \(1\), or one using \(p_{\mathbf{A}}\) produces a leading coefficient other than \((-1)^n\), recheck the determinant expansion. The check concerns the leading term only; it does not determine the other coefficients or the roots by itself.

Common Sign Mistakes

Mistake: treating the two conventions as equal in every dimension. In the three-dimensional example, \(q_{\mathbf{A}}(\lambda)=(\lambda-1)(\lambda-2)(\lambda-3)\), while \(p_{\mathbf{A}}(\lambda)=-(\lambda-1)(\lambda-2)(\lambda-3)\). Writing \(p_{\mathbf{A}}=q_{\mathbf{A}}\) gives the wrong value at \(\lambda=0\): the correct values are \(q_{\mathbf{A}}(0)=-6\) and \(p_{\mathbf{A}}(0)=6\). The correction is to count the dimension and apply the factor \((-1)^n\).

Mistake: changing the sign of the roots. A sign change in the polynomial does not mean that each root \(\lambda\) should be replaced by \(-\lambda\). For example, the two polynomials in the three-dimensional calculation vanish at \(1,2,3\), not at \(-1,-2,-3\). Multiplying a function by \(-1\) preserves its zeros because \(-f(\lambda)=0\) exactly when \(f(\lambda)=0\).

When comparing work from different sources, first identify which determinant is being used, then check the dimension. Translate the polynomial by multiplying the whole expression by \((-1)^n\); leave the parameter and its roots unchanged. That procedure accounts for the convention without changing the eigenvalue test.

Check Your Understanding

Use the determinant relation and the examples above to answer each question.

For an \(n\times n\) matrix with odd \(n\), how are \(p_{\mathbf{A}}\) and \(q_{\mathbf{A}}\) related?
For an even-dimensional matrix, do the two conventions have the same values or opposite values?
If \(q_{\mathbf{A}}(\lambda)=0\), what does the sign relation imply about \(p_{\mathbf{A}}(\lambda)\), and why?
In dimension \(3\), what is the leading coefficient of \(p_{\mathbf{A}}\)? What is the leading coefficient of \(q_{\mathbf{A}}\)?
For the three-dimensional worked example, check directly from the factored form whether \(\lambda=2\) is a root of each convention.

Characteristic Polynomial Sign Conventions

What You'll Learn

Two Conventions for the Same Eigenvalue Test