# The Unapologetic Mathematician

## The Characteristic Polynomial

Given a linear endomorphism $T:V\rightarrow V$ on a vector space $V$ of dimension $d$, we’ve defined a function$\det(T-\lambda1_V)$ — whose zeroes give exactly those field elements $\lambda$ so that the $\lambda$-eigenspace of $T$ is nontrivial. Actually, we’ll switch it up a bit and use the function $\det(\lambda1_V-T)$, which has the same useful property. Now let’s consider this function a little more deeply.

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

$\displaystyle\sum\limits_{\pi\in S_d}\mathrm{sgn}(\pi)\prod\limits_{k=1}^dt_k^{\pi(k)}$

which is a sum of products of matrix entries. Now, the matrix entries for the transformation $\lambda1_V-T$ are given by $t_i^j-\lambda\delta_i^j$. Each of these new entries is a polynomial (either constant or linear) in the variable $\lambda$. Any sum of products of polynomials is again a polynomial, and so our function is actually a polynomial in $\lambda$. We call it the “characteristic polynomial” of the transformation $T$. In terms of the matrix entries of $T$ itself, we get

$\displaystyle\sum\limits_{\pi\in S_d}\mathrm{sgn}(\pi)\prod\limits_{k=1}^d(\lambda\delta_k^{\pi(k)}-t_k^{\pi(k)})$

What’s the degree of this polynomial? Well, first let’s consider the degree of each term in the sum. Given a permutation $\pi\in S_d$ the term is the product of $d$ factors. The $k$th of these factors will be a field element if $k\neq\pi(k)$, and will be a linear polynomial if $k=\pi(k)$. Since multiplying polynomials adds their degrees, the degree of the $\pi$ term will be the number of $k$ such that $k=\pi(k)$. Thus the highest possible degree happens if $k=\pi(k)$ for all index values $k$. 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 $d$, equal to the dimension of the vector space $V$.

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

$\prod\limits_{k=1}^d(\lambda-t_k^k)$

Each factor gives $\lambda$ a coefficient of ${1}$, and so the coefficient of the $\lambda^d$ term is also ${1}$. Thus the leading coefficient of the characteristic polynomial is ${1}$ — a fact which turns out to be useful.

January 28, 2009 Posted by | Algebra, Linear Algebra | 13 Comments