Characteristic Polynomial

Topics: Algebra - Eigenvalues and Eigenvectors

(definition)

Let $A\in M_{n\times n}(\mathbb{K})$.

The polynomial $\det (A-\lambda I)=0$ is called the eigenpolynomial (polinomio característico).

(definition)

Let $T:V\to V$ and let $\beta$ be a basis for $V$. We define the eigenpolynomial $f(\lambda )$ of T as:

$$f(\lambda )=\det (A-\lambda I)=0$$

…where $A=[T{]}_{\beta }$. It’s common to find $\lambda$ denoted as $t$ in books and other documentation.

We solve this polynomial find the eigenvalues of a given linear transformation.

(note)

The eigenpolynomial is a polynomial with $n$ degree that has $(-1{)}^n$ as its main coefficient.

(theorem)

Let $A\in M_{n\times n}(\mathbb{K})$ and $T:V\to V$ with $V$ a vector space with finite dimension.

1. The scalar $\lambda$ is an eigenvalue of $A$ $⟺$:
   • $f(\lambda )=\det (A-\lambda I)=0$
   • $f(\lambda )=\det ([T{]}_{\beta }-\lambda I)=0$ ($\beta$ a basis for $V$)
2. $A$ has, at most, $n$ distinct eigenvalues; $T$ has, at most, $n$ distinct eigenvalues

Equivalences §

Let $T:V\to V$ with $V$ a vector space with $n$ finite dimension.

Let’s suppose that the characteristic polynomial of $T$ can be broken down into a product of degree 1 factors. Let ${\lambda }_1,\dots ,{\lambda }_k$ be distinct eigenvalues of $T$.

Then, the following propositions are equivalent:

1. $T$ is diagonalisable
2. $V=E_{{\lambda }_1}\oplus E_{{\lambda }_2}\oplus \cdots \oplus E_{{\lambda }_k}$
3. If $d_j=\dim (E_j)$ with $1\le j\le k$, then $d_1+\cdots +d_k=n$
4. If $m_j$ is the multiplicity of ${\lambda }_j$, then $\dim (E_{{\lambda }_j})=m_j$