Editing Cayley–Hamilton theorem (section)

=== {{math|2&thinsp;×&thinsp;2}} matrices ===

As a concrete example, let
<math display="block">A = \begin{pmatrix}1&2\\3&4\end{pmatrix}.</math>
Its characteristic polynomial is given by
<math display="block">
\begin{align}
p(\lambda) &= \det(\lambda I_2-A) = \det\!\begin{pmatrix}\lambda-1&-2\\-3&\lambda-4\end{pmatrix} \\
&=(\lambda-1)(\lambda-4)-(-2)(-3)=\lambda^2-5\lambda-2.
\end{align}
</math>

The Cayley–Hamilton theorem claims that, if we ''define''
<math display="block">p(X) = X^2 - 5X - 2I_2,</math>
then
<math display="block">p(A) = A^2-5A-2I_2 = \begin{pmatrix}0&0\\0&0\\\end{pmatrix}.</math>
We can verify by computation that indeed,
<math display="block">A^2-5A-2I_2 = \begin{pmatrix}7&10\\15&22\\\end{pmatrix} - \begin{pmatrix}5&10\\15&20\\\end{pmatrix} - \begin{pmatrix}2&0\\0&2\\\end{pmatrix} = \begin{pmatrix}0&0\\0&0\\\end{pmatrix}.</math>

For a generic {{math|2&thinsp;×&thinsp;2}} matrix,
<math display="block">A=\begin{pmatrix}a&b\\c&d\\\end{pmatrix} ,</math>

the characteristic polynomial is given by {{math|1= ''p''(''λ'') = ''λ''<sup>2</sup> − (''a'' + ''d'')''λ'' + (''ad'' − ''bc'')}}, so the Cayley–Hamilton theorem states that
<math display="block">p(A) = A^2-(a+d)A+(ad-bc)I_2 = \begin{pmatrix}0&0\\0&0\end{pmatrix};</math>
which is indeed always the case, evident by working out the entries of {{math|''A''<sup>2</sup>}}.

{{math proof|proof =
<math display="block">\begin{align}
&{} A^2-(a+d)A+(ad-bc)I_2 \\[1ex]
&= \begin{pmatrix}a^2+bc&ab+bd\\ac+cd&bc+d^2\\\end{pmatrix} - \begin{pmatrix}a(a+d)&b(a+d)\\c(a+d)&d(a+d) \end{pmatrix}+(ad-bc)I_2 \\[1ex]
&= \begin{pmatrix}bc-ad&0\\0&bc-ad\\\end{pmatrix}+(ad-bc)I_2 \\[1ex]
&= \begin{pmatrix}0&0\\0&0 \end{pmatrix}
\end{align}</math>
}}