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 \lambdaeigenspace 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


  1. […] 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 […]

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  2. […] 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 […]

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

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  4. […] 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 […]

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  5. […] 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 | Reply

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

    Comment by anju | February 13, 2009 | Reply

  7. A little demanding, aren’t we?

    Comment by John Armstrong | February 13, 2009 | Reply

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

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

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  10. […] 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 […]

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  11. […] 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 […]

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  12. […] 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 […]

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  13. […] 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 | Reply

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