# Linear mapping/Characteristic polynomial/Introduction/Section

We want to determine, for a given endomorphism , the eigenvalues and the eigenspaces. For this, the characteristic polynomial is decisive.

For an -matrix with entries in a field , the polynomial

is called the *characteristic polynomial*^{[1]}

For , this means

In this definition, we use the determinant of a matrix, which we have only defined for matrices with entries in a field. The entries are now elements of the polynomial ring . But, since we can consider these elements also inside the
field of rational functions
,^{[2]}
this is a useful definition. By definition, the determinant is an element in , but, because all entries of the matrix are polynomials, and because in the recursive definition of the determinant, only addition and multiplication is used, the characteristic polynomial is indeed a polynomial. The degree of the characteristic polynomial is , and its leading coefficient is , so it has the form

We have the important relation

for every , see exercise. Here, on the left-hand side, the number is inserted into the polynomial, and on the right-hand side, we have the determinant of a matrix which depends on .

For a linear mapping

on a finite-dimensional vector space, the *characteristic polynomial* is defined by

where is a describing matrix with respect to some basis. The multiplication theorem for the determinant shows that this definition is independent of the choice of the basis, see exercise.

Let denote a field, and let denote an -dimensional vector space. Let

denote a linear mapping. Then is an eigenvalue of , if and only if is a zero of the characteristic polynomial .

Let denote a describing matrix for , and let be given. We have

if and only if the linear mapping

is not bijective (and not injective) (due to fact and fact). This is, because of fact and fact, equivalent with

and this means that the eigenspace for is not the nullspace, thus is an eigenvalue for .

We consider the real matrix . The characteristic polynomial is

The eigenvalues are therefore (we have found these eigenvalues already in example, without using the characteristic polynomial).

For the matrix

the characteristic polynomial is

Finding the zeroes of this polynomial leads to the condition

which has no solution over , so that the matrix has no eigenvalues over . However, considered over the complex numbers , we have the two eigenvalues and . For the eigenspace for , we have to determine

a basis vector (hence an eigenvector) of this is . Analogously, we get

For an upper triangular matrix

the characteristic polynomial is

due to fact. In this case, we have directly a factorization of the characteristic polynomial into linear factors, so that we can see immediately the zeroes and the eigenvalues of , namely just the diagonal elements (which might not be all different).