characteristic polynomial

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 among its coefficients. The characteristic polynomial of an endomorphism of a finite-dimensional vector space is the characteristic polynomial of the matrix of that endomorphism over any base (that is, the characteristic polynomial does not depend on the choice of a basis). The characteristic equation, also known as the determinantal equation, is the equation obtained by equating the characteristic polynomial to zero.

In spectral graph theory, the characteristic polynomial of a graph is the characteristic polynomial of its adjacency matrix.

Motivation

In linear algebra, eigenvalues and eigenvectors play a fundamental role, since, given a linear transformation, an eigenvector is a vector whose direction is not changed by the transformation, and the corresponding eigenvalue is the measure of the resulting change of magnitude of the vector.

More precisely, if the transformation is represented by a square matrix $A,$ an eigenvector $\mathbf,$ and the corresponding eigenvalue $\lambda$ must satisfy the equation
$A \mathbf = \lambda \mathbf,$
or, equivalently,
$(\lambda I - A) \mathbf = 0$
where $I$ is the identity matrix, and $\mathbf\ne \mathbf$
(although the zero vector satisfies this equation for every $\lambda,$ it is not considered as an eigenvector).

It follows that the matrix $(\lambda I - A)$ must be singular, and its determinant
$\det(\lambda I - A) = 0$
must be zero.

In other words, the eigenvalues of are the roots of
$\det(xI - A),$
which is a monic polynomial in of degree if is a matrix. This polynomial is the ''characteristic polynomial'' of .

Formal definition

Consider an $n \times n$ matrix $A.$ The characteristic polynomial of $A,$ denoted by $p_A(t),$ is the polynomial defined by
$p_A(t) = \det (t I - A)$
where $I$ denotes the $n \times n$ identity matrix.

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 definition above always gives a monic polynomial, whereas the alternative definition is monic only when $n$ is even.

Examples

To compute the characteristic polynomial of the matrix
$A = \begin 2 & 1\\ -1& 0 \end.$
the determinant of the following is computed:
$t I-A = \begin t-2&-1\\ 1&t-0 \end$
and found to be $(t-2)t - 1(-1) = t^2-2t+1 \,\!,$ the characteristic polynomial of $A.$

Another example uses hyperbolic functions of a hyperbolic angle φ.
For the matrix take
$A = \begin \cosh(\varphi) & \sinh(\varphi)\\ \sinh(\varphi)& \cosh(\varphi) \end.$
Its characteristic polynomial is
$\det (tI - A) = (t - \cosh(\varphi))^2 - \sinh^2(\varphi) = t^2 - 2 t \ \cosh(\varphi) + 1 = (t - e^\varphi) (t - e^).$

Properties

The characteristic polynomial $p_A(t)$ of a $n \times n$ matrix 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 $p_A(t)$ (this also holds for the minimal polynomial of $A,$ but its degree may be less than $n$). All coefficients of the characteristic polynomial are polynomial expressions in the entries of the matrix. In particular its constant coefficient $p_A(0)$ is $\det(-A) = (-1)^n \det(A),$ the coefficient of $t^n$ is one, and the coefficient of $t^$ is , where is the trace of $A.$ (The signs given here correspond to the formal definition given in the previous section; for the alternative definition these would instead be $\det(A)$ and respectively.)

For a $2 \times 2$ matrix $A,$ the characteristic polynomial is thus given by
$t^2 - \operatorname(A) t + \det(A).$

Using the language of exterior algebra, the characteristic polynomial of an $n \times n$ matrix $A$ may be expressed as
$p_A (t) = \sum_^n t^ (-1)^k \operatorname\left(\textstyle\bigwedge^k A\right)$
where $\operatorname\left(\bigwedge^k A\right)$ is the trace of the $k$th exterior power of $A,$ which has dimension $\binom .$ This trace may be computed as the sum of all principal minors of $A$ of size $k.$ The recursive Faddeev–LeVerrier algorithm computes these coefficients more efficiently.

When the characteristic of the field of the coefficients is $0,$ each such trace may alternatively be computed as a single determinant, that of the $k \times k$ matrix,
$\operatorname\left(\textstyle\bigwedge^k A\right) = \frac \begin \operatornameA & k-1 &0&\cdots & \\ \operatornameA^2 &\operatornameA& k-2 &\cdots & \\ \vdots & \vdots & & \ddots & \vdots \\ \operatornameA^ &\operatornameA^& & \cdots & 1 \\ \operatornameA^k &\operatornameA^& & \cdots & \operatornameA \end ~.$

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 $n \times n$ matrices then characteristic polynomials of $AB$ and $BA$ coincide:
$p_(t)=p_(t).\,$

When $A$ is non-singular this result follows from the fact that $AB$ and $BA$ are similar:
$BA = A^ (AB) A.$

For the case where both $A$ and $B$ are singular, the desired identity is an equality between polynomials in $t$ and the coefficients of the matrices. Thus, to prove this equality, it suffices to prove that it is verified on a non-empty open subset (for the usual topology, or, more generally, for the Zariski topology) of the space of all the coefficients. As the non-singular matrices form such an open subset of the space of all matrices, this proves the result.

More generally, if $A$ is a matrix of order $m \times n$ and $B$ is a matrix of order $n \times m,$ then $AB$ is $m \times m$ and $BA$ is $n \times n$ matrix, and one has
$p_(t) = t^ p_(t).\,$

To prove this, one may suppose $n > m,$ by exchanging, if needed, $A$ and $B.$ Then, by bordering $A$ on the bottom by $n - m$ rows of zeros, and $B$ on the right, by, $n - m$ columns of zeros, one gets two $n \times n$ matrices $A^$ and $B^$ such that $B^A^ = BA$ and $A^B^$ is equal to $AB$ bordered by $n - m$ rows and columns of zeros. The result follows from the case of square matrices, by comparing the characteristic polynomials of $A^B^$ and $AB.$

Characteristic polynomial of ''A''^''k''

If $\lambda$ is an eigenvalue of a square matrix $A$ with eigenvector $\mathbf,$ then $\lambda^k$ is an eigenvalue of $A^k$ because
$A^k \textbf = A^ A \textbf = \lambda A^ \textbf = \dots = \lambda^k \textbf.$

The multiplicities can be shown to agree as well, and this generalizes to any polynomial in place of $x^k$:

That is, the algebraic multiplicity of $\lambda$ in $f(A)$ equals the sum of algebraic multiplicities of $\lambda'$ in $A$ over $\lambda'$ such that $f(\lambda') = \lambda.$
In particular, $\operatorname(f(A)) = \textstyle\sum_^n f(\lambda_i)$ and $\operatorname(f(A)) = \textstyle\prod_^n f(\lambda_i).$
Here a polynomial $f(t) = t^3+1,$ for example, is evaluated on a matrix $A$ simply as $f(A) = A^3+1.$

The theorem applies to matrices and polynomials over any field or commutative ring.
However, the assumption that $p_A(t)$ has a factorization into linear factors is not always true, unless the matrix is over an algebraically closed field such as the complex numbers.

Secular function and secular equation