MA20216: Algebra 2A

3.4 Eigenvalues and the characteristic polynomial

Recall from Chapter 3 of Algebra 1B:

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
\(\seteqnumber{0}{3.}{4}\)
\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
\(\seteqnumber{0}{3.}{4}\)
\begin{equation*} E_{\phi }(\lambda ):=\ker (\phi -\lambda \id _V)\leq V. \end{equation*}

Remark. Thus \(E_{\phi }(\lambda )\) consists of all eigenvectors of \(\phi \) with eigenvalue \(\lambda \) along with \(0\).

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
\(\seteqnumber{0}{3.}{4}\)
\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 \(\deg \Delta _{\phi }=\dim V\).

The characteristic polynomial is important to us because:

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

This prompts:

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 \(\dim E_{\phi }(\lambda )\).

From Algebra 1B¹,we know that \(\am (\lambda )\geq \gm (\lambda )\) and we will get a geometric understanding of \(\am (\lambda )\) in the next chapter (see §4.3.2).

When \(\F =\C \), Theorem 3.2, the Fundamental Theorem of Algebra, ensures that the characteristic polynomial has at least one root so we conclude from Lemma 3.8:

¹ Proposition 3.4.6.

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

Remark. This was crucial in Algebra 1B for the proof of the Spectral Theorem and will be equally crucial for us in the next chapter.

Eigenvalues and eigenvectors play nicely with polynomials:

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 \(\lambda \):
\(\seteqnumber{0}{3.}{4}\)
\begin{equation} \label {eq:8} \phi (v)=\lambda v. \end{equation}

Let \(p\in \F [x]\). Then
\(\seteqnumber{0}{3.}{5}\)
\begin{equation*} p(\phi )(v)=p(\lambda )v, \end{equation*}

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

Proof. The idea is to iterate (3.5):
\(\seteqnumber{0}{3.}{5}\)
\begin{equation*} \phi ^2(v)=\phi (\phi (v))=\phi (\lambda v)=\lambda \phi (v)=\lambda ^2v \end{equation*}

and so, by induction, \(\phi ^k(v)=\lambda ^kv\), for all \(k\in \N \).

Now, for \(p=\sum _{k=0}^{n}a_{k}x^k\),
\(\seteqnumber{0}{3.}{5}\)
\begin{equation*} p(\phi )(v)=\sum _{k=0}^{n}a_k\phi ^k(v)=\sum _{k=0}^{n}a_k\lambda ^kv =\bigl (\sum _{k=0}^{n}a_k\lambda ^k\bigr )v=p(\lambda )v. \end{equation*}

□

This gives us something interesting: if \(p(\phi )=0\) then

\begin{equation*} 0=p(\phi )(v)=p(\lambda )v \end{equation*}

so that, since \(v\neq 0\), \(p(\lambda )=0\). Thus any eigenvalue of \(\phi \) is a root of \(p\). In particular:

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 \(m_{\phi }\).