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Characteristic polynomial

by , PhD

The characteristic polynomial of a square matrix is the polynomial that has the eigenvalues of the matrix as its roots.

We have already introduced the characteristic polynomial in the lecture on eigenvalues. Here we study its properties in greater detail.

Table of Contents


Here is a definition.

Definition Let A be a $K	imes K$ matrix. The characteristic polynomial of A is the polynomial[eq1]where I is the $K	imes K$ identity matrix.

Here is a simple example.

Example Define the $2	imes 2$ matrix[eq2]Then[eq3]Therefore, the characteristic polynomial of A is[eq4]


The characteristic polynomial is monic (i.e., the coefficient of its highest power is 1) and its degree is equal to the dimension of the matrix.

Proposition Let A be a $K	imes K$ matrix. The characteristic polynomial of A is a monic polynomial of degree K.


This proposition can be proved by using the definition of determinant[eq5]where $P$ is the set of all permutations of the first K natural numbers. Thus, $cleft( z
ight) $ is a sum of polynomials of the form[eq6]The polynomial of this form having the highest degree is that in which all the factors [eq7] are diagonal elements of $zI-A$. It corresponds to the permutation $pi _{0}$ in which the K natural numbers are sorted in increasing order. The parity of $pi _{0}$ is even and its sign is [eq8]because it does not contain any inversion (see the lecture on the sign of a permutation). Thus, the summand having the highest degree is[eq9]which has degree K and is monic. All the other summands have degree less than K. Therefore, $c$ has degree K and is monic.


Being a monic polynomial of degree K, the characteristic polynomial can be written as[eq10]

Hence,[eq11]where the last equality is a consequence of the properties of the determinant.


We also have the following property:[eq12]where [eq13] is the trace of A. The proof of this fact can be found in a solved exercise at the end of this lecture.

Fundamental theorem of algebra

By the Fundamental Theorem of Algebra, a monic polynomial of degree K whose coefficients are complex can be factored into the product of K linear factors (revise the lecture on polynomials if you are puzzled). As a consequence, the characteristic polynomial can be written as[eq14]where [eq15] are the roots of $cleft( z
ight) $, that is, the values such that[eq16]

In other words, the roots of the characteristic polynomial are the eigenvalues of A.

Solved exercises

Below you can find some exercises with explained solutions.

Exercise 1

Prove the above claim that[eq12]


We have seen above that the characteristic polynomial $cleft( z
ight) $ is a sum of polynomials:[eq18]where $pi $ is a permutation of the first K natural numbers. We have already seen that there is only one summand that contains a $z^{K}$ term, corresponding to the permutation such that [eq19]. This is also the only summand that contains a $z^{K-1}$ term because, as soon as we invert the order of two numbers in the permutation, two diagonal terms drop out of the product [eq20]Hence, we just need to find the coefficient of $z^{K-1}$ in the product [eq21]By expanding the product, we can see that all the $z^{K-1}$ terms are of the form [eq22]and there are K such terms (for $j=1,ldots ,K$). Therefore, [eq23]

How to cite

Please cite as:

Taboga, Marco (2017). "Characteristic polynomial", Lectures on matrix algebra.

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