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3.5 The Cayley–Hamilton theorem
-
Theorem 3.12 (Cayley–Hamilton4 Theorem). Let \(\phi \in
L(V)\) be a linear operator on a finite-dimensional vector space over a field \(\F \).
Then \(\Delta _{\phi }(\phi )=0\).
Equivalently, for any \(A\in M_n(\F )\), \(\Delta _A(A)=0\).
Before proving this, let us see what it tells us. Let
\(\seteqnumber{0}{3.}{6}\)
\begin{equation*}
A= \begin{pmatrix} a&b\\c&d \end {pmatrix}\in M_2(\F ).
\end{equation*}
Then
\(\seteqnumber{0}{3.}{6}\)
\begin{equation*}
\Delta _A= \begin{vmatrix} a-x&b\\c&d-x \end {vmatrix}=x^2-(a+d)x+(ad-bc).
\end{equation*}
So the Cayley–Hamilton theorem is telling us that
\(\seteqnumber{0}{3.}{6}\)
\begin{equation*}
A^2-(a+d)A+(ad-bc)I_2=0,
\end{equation*}
that is,
\(\seteqnumber{0}{3.}{6}\)
\begin{equation*}
\begin{pmatrix} a^2+bc&ab+bd\\ca+dc&cb+d^2 \end {pmatrix} -(a+d) \begin{pmatrix} a&b\\c&d \end {pmatrix} + \begin{pmatrix} ad-bc&0\\0&ad-bc \end {pmatrix} = \begin{pmatrix} 0&0\\0&0 \end {pmatrix}.
\end{equation*}
This is certainly true (check it!) but is far from obvious! If you are not yet convinced, work out what the theorem says for \(A\in M_3(\F )\).
-
Proof of Theorem 3.12. We will prove the matrix version. So let \(A\in M_n(\F )\) and write
\(\seteqnumber{0}{3.}{6}\)
\begin{equation*}
\Delta _{A}=a_0+\dots +a_nx^n.
\end{equation*}
Thus, our mission is to show that
\(\seteqnumber{0}{3.}{6}\)
\begin{equation*}
a_0I_n+a_1A+\dots +a_nA^n=0.
\end{equation*}
The key is the adjugate formula from Algebra 1B5:
\(\seteqnumber{0}{3.}{6}\)
\begin{equation}
\label {eq:11} \mathrm {adj}(A-xI_{n})(A-xI_{n})=\det (A-xI_{n})I_{n}.
\end{equation}
Each entry of \(\mathrm {adj}(A-xI_n)\) is a polynomial in \(x\) of degree at most \(n-1\) so we write
\(\seteqnumber{0}{3.}{7}\)
\begin{equation*}
\mathrm {adj}(A-xI_n)=B_0+B_1x+\dots +B_{n-1}x^{n-1},
\end{equation*}
with each \(B_k\in M_n(\F )\). Substitute this into (3.7) to get
\(\seteqnumber{0}{3.}{7}\)
\begin{equation*}
(B_0+B_1x+\dots +B_{n-1}x^{n-1})(A-xI)=(a_0+\dots +a_nx^n)I_n
\end{equation*}
and compare coefficients of \(x^{k}\) to get
\(\seteqnumber{0}{3.}{7}\)
\begin{equation}
\label {eq:12} B_kA-B_{k-1}=a_kI_{n},
\end{equation}
for \(0\leq k\leq n\), where we have set \(B_{-1}=B_n=0\in M_n(\F )\).
Multiply (3.8) by \(A^k\) on the right to get
\(\seteqnumber{0}{3.}{8}\)
\begin{equation*}
B_kA^{k+1}-B_{k-1}A^k=a_kA^k
\end{equation*}
and sum:
\(\seteqnumber{0}{3.}{8}\)
\begin{equation*}
\Delta _{A}(A)=\sum _{k=0}^na_kA^k=\sum _{k=0}^n(B_kA^{k+1}-B_{k-1}A^k)=B_nA^{n+1}-B_{-1}=0
\end{equation*}
because nearly all terms in the penultimate sum cancel. □
-
Proof. By Theorem 3.12, \(\Delta _{\phi }(\phi )=0\) so \(m_{\phi }\) divides \(\Delta _{\phi }\) by Theorem 3.7. As a result, any root of \(m_{\phi }\) is a root of \(\Delta _{\phi }\) and so an eigenvalue. Conversely, any eigenvalue is a root of \(m_{\phi }\) by Theorem 3.11. □
Let us summarise the situation when \(\F =\C \) so that any polynomial is a product of linear factors. So let \(\phi \in L(V)\) be a linear operator on a finite-dimensional complex vector space with distinct eigenvalues \(\lst \lambda 1k\). Then
\(\seteqnumber{0}{3.}{8}\)
\begin{align*}
\Delta _{\phi }&=\pm \prod _{i=1}^k(x-\lambda _i)^{r_i}\\ m_{\phi }&=\prod _{i=1}^k(x-\lambda _i)^{s_i},
\end{align*}
where \(r_i=\am (\lambda _i)\) and \(\bw 1{s_i}{r_i}\), for \(\bw 1ik\).
This gives us another way to find \(m_{\phi }\) if we can factorise \(\Delta _{\phi }\): \(m_{\phi }\) will be of the form \(p=\prod _{i=1}^{k}(x-\lambda _{i})^{s_i}\), with each \(\bw 1{s_i}{r_i}\), so evaluate \(p(\phi )\) to find the one of lowest degree with
\(p(\phi )=0\).