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.

Here is a definition.

Definition Let be a matrix. The characteristic polynomial of is the polynomialwhere is the identity matrix.

Here is a simple example.

Example Define the matrixThenTherefore, the characteristic polynomial of is

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

Proposition Let be a matrix. The characteristic polynomial of is a monic polynomial of degree .

Proof

This proposition can be proved by using the definition of determinantwhere is the set of all permutations of the first natural numbers. Thus, is a sum of polynomials of the formThe polynomial of this form having the highest degree is that in which all the factors are diagonal elements of . It corresponds to the permutation in which the natural numbers are sorted in increasing order. The parity of is even and its sign is because it does not contain any inversion (see the lecture on the sign of a permutation). Thus, the summand having the highest degree iswhich has degree and is monic. All the other summands have degree less than . Therefore, has degree and is monic.

Being a monic polynomial of degree , the characteristic polynomial can be written as

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

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

By the Fundamental Theorem of Algebra, a monic polynomial of degree whose coefficients are complex can be factored into the product of linear factors (revise the lecture on polynomials if you are puzzled). As a consequence, the characteristic polynomial can be written aswhere are the roots of , that is, the values such that

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

Below you can find some exercises with explained solutions.

Prove the above claim that

Solution

We have seen above that the characteristic polynomial is a sum of polynomials:where is a permutation of the first natural numbers. We have already seen that there is only one summand that contains a term, corresponding to the permutation such that . This is also the only summand that contains a term because, as soon as we invert the order of two numbers in the permutation, two diagonal terms drop out of the product Hence, we just need to find the coefficient of in the product By expanding the product, we can see that all the terms are of the form and there are such terms (for ). Therefore,

Please cite as:

Taboga, Marco (2017). "Characteristic polynomial", Lectures on matrix algebra. https://www.statlect.com/matrix-algebra/characteristic-polynomial.

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