The Unapologetic Mathematician

Mathematics for the interested outsider

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 kth 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


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



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