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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
\(\seteqnumber{0}{3.}{5}\)
\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
\(\seteqnumber{0}{3.}{5}\)
\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
\(\seteqnumber{0}{3.}{5}\)
\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:
This prompts:
From Algebra 1B3,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 Theorem 3.8:
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 \(\lambda \):
\(\seteqnumber{0}{3.}{5}\)
\begin{equation}
\label {eq:10} \phi (v)=\lambda v.
\end{equation}
Let \(p\in \F [x]\). Then
\(\seteqnumber{0}{3.}{6}\)
\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 (3.6):
\(\seteqnumber{0}{3.}{6}\)
\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 \(p=\sum _{k=0}^{n}a_{k}x^k\),
\(\seteqnumber{0}{3.}{6}\)
\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
\(\seteqnumber{0}{3.}{6}\)
\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: