Cayley-Hamilton Theorem

Theorem

Let \(F\) be a field, and let \(V\) be a finite dimensional \(F\)-vector space. If \(t : V → V\) is a linear transformation, then \(m_t\mid c_t \), and hence \(c_t(t) = 0\). Similarly, for any matrix \(A ∈ M_n(F)\) over a field \(F\) we have \(m_A|c_A\) and \(c_A(A) = 0\).

Proof

Let \( A = [t]_B^B\) for some basis \(B\) of \(V\). Note that the statements about \(A\) and \(t\) are equivalent, since by definition \(c_A = c_t\) , while \(m_A = m_t\) we have \(f(A) = 0\) if and only if \(f(t) = 0\). So write \(m = m_A = m_t\) and \(c = c_A = c_t\). By Lemma 4.21, \(m = g_k\) and \(c = g_1 · · · g_k\), so \(m | c\). By definition, we \(m(A) = 0\). Since \(m|c\), we conclude that \(c(A) = 0\). \( \)

Lemma 4.21

Let \(F \) be a field, let \( V\) be a finite dimensional \( F\)-vector space, and \( _V → V\) be a linear transformation with invariant factors \(g_1| · · · |g_k. \) Then \( c_t = g_1 · · · g_k\) and mt = gk.

Proof

The product of the elements on the diagonal of the Smith Normal Form of \( xI_n − A\) is the determinant of \(xI_n − A\). Thus the product of the invariant factors \(g_1 · · · g_k\) of \(V_t\) is the characteristic polynomial \(c_t\) of \(t\). Notice here that we chose our invariant factors \(g_1, . . . , g_k\) to be monic, so that \(g_1 · · · g_k\) is monic, and thus actually equal to \(c_t\) (not just up to multiplication by a unit). By Problem Set 5, \(\text{ann}_{F[x]}(V_t) = (g_k)\), and since \(g_k\) is monic we deduce that \(m_t = g_k\).