# Characteristic polynomial

This article is about the characteristic polynomial of a matrix or of an endomorphism of vector spaces. For the characteristic polynomial of a matroid, see Matroid. For that of a graded poset, see Graded poset.

In linear algebra, the characteristic polynomial of a square matrix is a polynomial, which is invariant under matrix similarity and has the eigenvalues as roots. It has the determinant and the trace of the matrix as coefficients. The characteristic polynomial of an endomorphism of vector spaces of finite dimension is the characteristic polynomial of the matrix of the endomorphism over any base; it does not depend on the choice of a basis. The characteristic equation is the equation obtained by equating to zero the characteristic polynomial.

The characteristic polynomial of a graph is the characteristic polynomial of its adjacency matrix. It is a graph invariant, though it is not complete: the smallest pair of non-isomorphic graphs with the same characteristic polynomial have five nodes.[1]

## Motivation

Given a square matrix A, we want to find a polynomial whose zeros are the eigenvalues of A. For a diagonal matrix A, the characteristic polynomial is easy to define: if the diagonal entries are a1a2a3, etc. then the characteristic polynomial will be:

$(t-a_1)(t-a_2)(t-a_3)\cdots.\,$

This works because the diagonal entries are also the eigenvalues of this matrix.

For a general matrix A, one can proceed as follows. A scalar λ is an eigenvalue of A if and only if there is an eigenvector v ≠ 0 such that

$A \mathbf{v} = \lambda \mathbf{v},\,$

or

$(\lambda I - A)\mathbf{v} = 0\,$

(where I is the identity matrix). Since v is non-zero, this means that the matrix λ I − A is singular (non-invertible), which in turn means that its determinant is 0. Thus the roots of the function det(λ I − A) are the eigenvalues of A, and it is clear that this determinant is a polynomial in λ.

## Formal definition

We start with a field K (such as the real or complex numbers) and an n×n matrix A over K. The characteristic polynomial of A, denoted by pA(t), is the polynomial defined by

$p_A(t) = \det \left(t \boldsymbol{I} - A\right)$

where I denotes the n-by-n identity matrix and the determinant is being taken in K[t], the ring of polynomials in t over K.

Some authors define the characteristic polynomial to be det(A - t I). That polynomial differs from the one defined here by a sign (−1)n, so it makes no difference for properties like having as roots the eigenvalues of A; however the current definition always gives a monic polynomial, whereas the alternative definition always has constant term det(A).

## Examples

Suppose we want to compute the characteristic polynomial of the matrix

$A=\begin{pmatrix} 2 & 1\\ -1& 0 \end{pmatrix}.$

We now compute the determinant of

$t I-A = \begin{pmatrix} t-2&-1\\ 1&t-0 \end{pmatrix}$ which is $(t-2)t - 1(-1) = t^2-2t+1 \,\!,$ the characteristic polynomial of A.

Another example uses hyperbolic functions of a hyperbolic angle φ. For the matrix take

$A=\begin{pmatrix} \cosh(\phi) & \sinh(\phi)\\ \sinh(\phi)& \cosh(\phi) \end{pmatrix}.$

Its characteristic polynomial is

$\det (tI - A) = (t - \cosh(\phi))^2 - \sinh^2(\phi) = t^2 - 2 t \ \cosh(\phi) + 1 = (t - e^\phi) (t - e^{-\phi}).$

## Properties

The polynomial pA(t) is monic (its leading coefficient is 1) and its degree is n. The most important fact about the characteristic polynomial was already mentioned in the motivational paragraph: the eigenvalues of A are precisely the roots of pA(t) (this also holds for the minimal polynomial of A, but its degree may be less than n). The coefficients of the characteristic polynomial are all polynomial expressions in the entries of the matrix. In particular its constant coefficient pA (0)  is det(−A) = (−1)n det(A), the coefficient of tn is one, and the coefficient of tn−1 is tr(−A) = −tr(A), where tr(A) is the matrix trace of A. (The signs given here correspond to the formal definition given in the previous section;[2] for the alternative definition these would instead be det(A) and (−1)n − 1 tr(A) respectively.[3])

For a 2×2 matrix A, the characteristic polynomial is thus given by

$t^2 - \operatorname{tr}(A) t + \det(A).$

Using the language of exterior algebra, one may compactly express the characteristic polynomial of an n×n matrix A as

$p_A (t) = \sum_{k=0}^n t^{n-k} (-1)^k \operatorname{tr}(\Lambda^k A)$

where tr(ΛkA) is the trace of the kth exterior power of A, with dimension $\tbinom nk$, and may be evaluated explicitly as the determinant of the k×k matrix,

$\frac{1}{k!} \begin{vmatrix} \operatorname{tr}A & k-1 &0&\cdots & \\ \operatorname{tr}A^2 &\operatorname{tr}A& k-2 &\cdots & \\ \vdots & \vdots & & \ddots & \vdots \\ \operatorname{tr}A^{k-1} &\operatorname{tr}A^{k-2}& & \cdots & 1 \\ \operatorname{tr}A^k &\operatorname{tr}A^{k-1}& & \cdots & \operatorname{tr}A \end{vmatrix} ~.$

The Cayley–Hamilton theorem states that replacing t by A in the characteristic polynomial (interpreting the resulting powers as matrix powers, and the constant term c as c times the identity matrix) yields the zero matrix. Informally speaking, every matrix satisfies its own characteristic equation. This statement is equivalent to saying that the minimal polynomial of A divides the characteristic polynomial of A.

Two similar matrices have the same characteristic polynomial. The converse however is not true in general: two matrices with the same characteristic polynomial need not be similar.

The matrix A and its transpose have the same characteristic polynomial. A is similar to a triangular matrix if and only if its characteristic polynomial can be completely factored into linear factors over K (the same is true with the minimal polynomial instead of the characteristic polynomial). In this case A is similar to a matrix in Jordan normal form.

## Characteristic polynomial of a product of two matrices

If A and B are two square n×n matrices then characteristic polynomials of AB and BA coincide:

$p_{AB}(t)=p_{BA}(t).\,$

More generally, if A is a matrix of order m×n and B is a matrix of order n×m, then AB is m×m and BA is n×n matrix. One has

$p_{BA}(t) = t^{n-m} p_{AB}(t).\,$

To prove the first result, recognize that the equation to be proved, as a polynomial in t and in the entries of A and B is a universal polynomial identity. It therefore suffices to check it on an open set of parameter values in the complex numbers. The tuples (A,B,t) where A is an invertible complex n by n matrix, B is any complex n by n matrix, and t is any complex number from an open set in complex space of dimension 2n2 + 1. When A is non-singular our result follows from the fact that AB and BA are similar:

$BA = A^{-1} (AB) A.\,$

## Secular function and secular equation

### Secular function

The terms secular function has been used for what is now called characteristic polynomial (in some literature the term secular function is still used). The term comes from the fact that the characteristic polynomial was used to calculate secular perturbations (on a time scale of a century, i.e. slow compared to annual motion) of planetary orbits, according to Lagrange's theory of oscillations.

### Secular equation

Secular equation may have several meanings.

• In linear algebra it is sometimes used in place of characteristic equation.
• In astronomy it is the algebraic or numerical expression of the magnitude of the inequalities in a planet's motion that remain after the inequalities of a short period have been allowed for.[4]
• In molecular orbital calculations relating to the energy of the electron and its wave function it is also used instead of the characteristic equation.