Difference between revisions of "Cayley-Hamilton Theorem"
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| + | ==Theorem== | ||
Let <math>F</math> be a field, and let <math>V</math> be a finite dimensional <math>F</math>-vector space. If <math>t : V → V</math> is a linear transformation, then <math>m_t\mid c_t | Let <math>F</math> be a field, and let <math>V</math> be a finite dimensional <math>F</math>-vector space. If <math>t : V → V</math> is a linear transformation, then <math>m_t\mid c_t | ||
</math>, and hence <math>c_t(t) = 0</math>. | </math>, and hence <math>c_t(t) = 0</math>. | ||
Similarly, for any matrix <math>A ∈ M_n(F)</math> over a field <math>F</math> we have <math>m_A|c_A</math> and <math>c_A(A) = 0</math>. | Similarly, for any matrix <math>A ∈ M_n(F)</math> over a field <math>F</math> we have <math>m_A|c_A</math> and <math>c_A(A) = 0</math>. | ||
| + | |||
| + | ==Proof== | ||
| + | Let <math> A = [t]_B^B</math> for some basis <math>B</math> of <math>V</math>. Note that the statements about <math>A</math> and <math>t</math> are equivalent, since by definition <math>c_A = c_t</math> , while <math>m_A = m_t</math> we have <math>f(A) = 0</math> if and only if <math>f(t) = 0</math>. So write <math>m = m_A = m_t</math> and <math>c = c_A = c_t</math>. | ||
| + | By Lemma 4.21, <math>m = g_k</math> and <math>c = g_1 · · · g_k</math>, so <math>m | c</math>. By definition, we <math>m(A) = 0</math>. Since <math>m|c</math>, we conclude that <math>c(A) = 0</math>. | ||
<math> </math> | <math> </math> | ||
Revision as of 18:12, 10 March 2023
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 \(c(A) = 0\). \( \)
