# 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.

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January 28, 2009 - Posted by | Algebra, Linear Algebra

## 13 Comments »

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 […]

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

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 […]

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

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 […]

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

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 […]

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

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 […]

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

9. […] 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 | Reply

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 […]

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

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 […]

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

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 […]

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

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