characteristic polynomial

Definition

The characteristic polynomial of a linear map $T:V\to V$ for a finite dimensional vector space $V$ over a field $k$ is

$${\chi }_T(x)=\det (x\cdot {\text{id}}_V-T)\in k[x]$$

Note that ${\chi }_T$ is a monic polynomial of degree equal to the dimension of $V$ whose roots are the eigenvalues of $T$.

Relation to rational canonical form

Let $T:V\to V$ be an endomorphism of a finite dimensional vector space over $k$ with rational canonical form

$$\left[\begin{array}{ccc}\begin{array}{c}{C}_{q_1}\end{array}& & 0\\ & \ddots & \\ 0& & \begin{array}{c}{C}_{q_K}\end{array}\end{array}\right]$$

Then ${\chi }_T=q_1\cdots q_K$.

Proof

We don’t have to do the whole matrix, it suffices to prove that ${\chi }_{C_q}=q$. Write $q$ as

$$q=x^d+a_{d-1}{x}^{d-1}+\cdots +a_0$$

so that

$$C_q=\left[\begin{array}{cccccc}0& \cdots & & \cdots & 0& -a_0\\ 1& \ddots & & & ⋮& ⋮\\ 0& \ddots & & \ddots & ⋮& ⋮\\ ⋮& \ddots & & \ddots & 0& -a_{d-2}\\ 0& \cdots & & 0& 1& -a_{d-1}\end{array}\right].$$

We compute $\det C_q$ by induction on degree of $q$. The base case ($d=1$) is

$${\chi }_{C_q}(x)=\det (x+a_0)=x+a_0=q$$

so we are good there.

In general, we expand along the first row to get

$$\begin{array}{rl}{\chi }_{C_q}(x)& =\det \left[\begin{array}{ccccc}x& 0& \cdots & 0& a_0\\ -1& \ddots & & ⋮& ⋮\\ 0& \ddots & \ddots & 0& ⋮\\ ⋮& \ddots & \ddots & x& a_{d-2}\\ 0& \cdots & 0& -1& x+a_{d-1}\end{array}\right]\\ & =x\det \left[\begin{array}{ccccc}x& 0& \cdots & 0& a_1\\ -1& \ddots & & ⋮& ⋮\\ 0& \ddots & \ddots & 0& ⋮\\ ⋮& \ddots & \ddots & x& a_{d-2}\\ 0& \cdots & 0& -1& x+a_{d-1}\end{array}\right]+(-1{)}^{d+1}{a}_0\det \left[\begin{array}{ccccc}-1& x& 0& 0& \\ 0& \ddots & \ddots & 0& \\ ⋮& \ddots & \ddots & x& \\ 0& \cdots & 0& -1& \end{array}\right]\\ & =x(x^{d-1}+a_{d-1}{x}^{d-2}+\cdots +a_1)+(-1{)}^{d-1}{a}_0(-1{)}^{d-1}\\ & =q\end{array}$$