MA20216: Algebra 2A

3.4 Eigenvalues and the characteristic polynomial

Definitions. Let \(V\) be a vector space over \(\F \) and \(\phi \in L(V)\).

An eigenvalue of \(\phi \) is a scalar \(\lambda \in \F \) such that there is a non-zero \(v\in V\) with
\(\seteqnumber{0}{3.}{2}\)
\begin{equation*} \phi (v)=\lambda v. \end{equation*}

Such a vector \(v\) is called an eigenvector of \(\phi \) with eigenvalue \(\lambda \).

The \(\lambda \)-eigenspace \(E_{\phi }(\lambda )\) of \(\phi \) is given by
\(\seteqnumber{0}{3.}{2}\)
\begin{equation*} E_{\phi }(\lambda ):=\ker (\phi -\lambda \id _V)\leq V. \end{equation*}

Definition. Let \(V\) be a finite-dimensional vector space over \(\F \) and \(\phi \in L(V)\).

The characteristic polynomial \(\Delta _{\phi }\) of \(\phi \) is given by
\(\seteqnumber{0}{3.}{2}\)
\begin{equation*} \Delta _{\phi }(\lambda ):=\det (\phi -\lambda \id _V)=\det (A-\lambda \I ), \end{equation*}

where \(A\) is the matrix of \(\phi \) with respect to some (any!) basis of \(V\).

Thus \(\deg \Delta _{\phi }=\dim V\).

Lemma 3.8. A scalar \(\lambda \in \F \) is an eigenvalue of \(\phi \) if and only if \(\lambda \) is a root of \(\Delta _{\phi }\).

Definitions. Let \(\phi \in L(V)\) be in a linear operator on a finite-dimensional vector space \(V\) over \(\F \) and \(\lambda \) an eigenvalue of \(\phi \). Then
- (1) The algebraic multiplicity of \(\lambda \), \(\am (\lambda )\in \Z _+\), is the multiplicity of \(\lambda \) as a root of \(\Delta _{\phi }\).
- (2) The geometric multiplicity of \(\lambda \), \(\gm (\lambda )\in \Z _+\), is \(\dim E_{\phi }(\lambda )\).

Theorem 3.9. Let \(\phi \) be a linear operator on a finite-dimensional vector space \(V\) over \(\C \). Then \(\phi \) has an eigenvalue.

Proposition 3.10. Let \(\phi \in L(V)\) be a linear operator on a vector space over a field \(\F \) and let \(v\in V\) be an eigenvector of \(\phi \) with eigenvalue \(\lambda \):
\(\seteqnumber{0}{3.}{2}\)
\begin{equation} \label {eq:8} \phi (v)=\lambda v. \end{equation}

Let \(p\in \F [x]\). Then
\(\seteqnumber{0}{3.}{3}\)
\begin{equation*} p(\phi )(v)=p(\lambda )v, \end{equation*}

so that \(v\) is an eigenvector of \(p(\phi )\) also with eigenvalue \(p(\lambda )\).

Corollary 3.11. Let \(\phi \) be a linear operator on a finite-dimensional vector space \(V\) over \(\F \). Then any eigenvalue of \(\phi \) is a root of \(m_{\phi }\).