## The Characteristic Polynomial

Given a linear endomorphism on a vector space of dimension , we’ve defined a function — — whose zeroes give exactly those field elements so that the -eigenspace of is nontrivial. Actually, we’ll switch it up a bit and use the function , which has the same useful property. Now let’s consider this function a little more deeply.

First off, if we choose a basis for we have matrix representations of endomorphisms, and thus a formula for their determinants. For instance, if is represented by the matrix with entries , then its determinant is given by

which is a sum of products of matrix entries. Now, the matrix entries for the transformation are given by . Each of these new entries is a polynomial (either constant or linear) in the variable . Any sum of products of polynomials is again a polynomial, and so our function is actually a polynomial in . We call it the “characteristic polynomial” of the transformation . In terms of the matrix entries of itself, we get

What’s the degree of this polynomial? Well, first let’s consider the degree of each term in the sum. Given a permutation the term is the product of factors. The th of these factors will be a field element if , and will be a linear polynomial if . Since multiplying polynomials adds their degrees, the degree of the term will be the number of such that . Thus the highest possible degree happens if for all index values . This only happens for one permutation — the identity — so there can’t be another term of the same degree to cancel the highest-degree monomial when we add them up. And so the characteristic polynomial has degree , equal to the dimension of the vector space .

What’s the leading coefficient? Again, the degree- monomial can only show up once, in the term corresponding to the identity permutation. Specifically, this term is

Each factor gives a coefficient of , and so the coefficient of the term is also . Thus the leading coefficient of the characteristic polynomial is — a fact which turns out to be useful.

[...] from the Characteristic Polynomial This one’s a pretty easy entry. If we know the characteristic polynomial of an endomorphism on a vector space of finite dimension , then we can get its determinant from [...]

Pingback by Extracting the Determinant from the Characteristic Polynomial « The Unapologetic Mathematician | January 29, 2009 |

[...] let’s take a linear endomorphism on a vector space of finite dimension . We know that its characteristic polynomial can be defined without reference to a basis of , and so each of the coefficients of is independent [...]

Pingback by The Trace of a Linear Transformation « The Unapologetic Mathematician | January 30, 2009 |

[...] does this assumption buy us? It says that the characteristic polynomial of a linear transformation is — like any polynomial over an algebraically closed field [...]

Pingback by Upper-Triangular Matrices « The Unapologetic Mathematician | February 2, 2009 |

[...] remember that we can always calculate the characteristic polynomial of . This will be a polynomial of degree — the dimension of . Further, we know that a field [...]

Pingback by Transformations with All Eigenvalues Distinct « The Unapologetic Mathematician | February 10, 2009 |

[...] So we’ve got a linear transformation on a vector space of finite dimension . We take its characteristic polynomial to find the eigenvalues. If all of its roots are distinct (and there are of them, as there must be [...]

Pingback by Repeated Eigenvalues « The Unapologetic Mathematician | February 11, 2009 |

please send me the coefficient in the characteristic polynomial of signed graph of order n.

Comment by anju | February 13, 2009 |

A little demanding, aren’t we?

Comment by John Armstrong | February 13, 2009 |

[...] we saw that if the entries along the diagonal of an upper-triangular matrix are , then the characteristic polynomial [...]

Pingback by The Multiplicity of an Eigenvalue « The Unapologetic Mathematician | February 19, 2009 |

[...] we can calculate the characteristic polynomial of , whose roots are the eigenvalues of . For each eigenvalue , we can define the generalized [...]

Pingback by Jordan Normal Form « The Unapologetic Mathematician | March 4, 2009 |

[...] of an Eigenpair An eigenvalue of a linear transformation is the same thing as a root of the characteristic polynomial of . That is, the characteristic polynomial has a factor . We can evaluate this polynomial at to [...]

Pingback by Eigenvectors of an Eigenpair « The Unapologetic Mathematician | April 3, 2009 |

[...] we found this in the complex case we saw that the characteristic polynomial had to have a root, since is algebraically closed. It’s the fact that isn’t [...]

Pingback by Every Self-Adjoint Transformation has an Eigenvector « The Unapologetic Mathematician | August 12, 2009 |

[...] on the remaining -dimensional space. This tells us that all of our eigenvalues are , and the characteristic polynomial is , where . We can evaluate this on the transformation to find [...]

Pingback by A Lemma on Reflections « The Unapologetic Mathematician | January 19, 2010 |

[...] the Jordan-Chevalley decomposition. We let have distinct eigenvalues with multiplicities , so the characteristic polynomial of [...]

Pingback by The Jordan-Chevalley Decomposition (proof) « The Unapologetic Mathematician | August 28, 2012 |