Editing Cayley–Hamilton theorem (section)

=== A bogus "proof": {{math|1=''p''(''A'') = det(''AI''<sub>''n''</sub> − ''A'') = det(''A'' − ''A'') = 0}} ===
One persistent elementary but ''incorrect'' argument<ref>{{harvnb|Garrett|2007|p=381}}</ref> for the theorem is to "simply" take the definition
<math display="block">p(\lambda) = \det(\lambda I_n - A)</math>
and substitute {{mvar|A}} for {{mvar|λ}}, obtaining
<math display="block">p(A)=\det(A I_n - A) = \det(A - A) = \det(\mathbf{0}) = 0.</math>

There are many ways to see why this argument is wrong. First, in the Cayley–Hamilton theorem, {{math|''p''(''A'')}} is an ''{{math|n&thinsp;×&thinsp;n}} matrix''. However, the right hand side of the above equation is the value of a determinant, which is a ''scalar''. So they cannot be equated unless {{math|1=''n'' = 1}} (i.e. {{mvar|A}} is just a scalar). Second, in the expression <math>\det(\lambda I_n - A)</math>, the variable λ actually occurs at the diagonal entries of the matrix <math>\lambda I_n - A</math>. To illustrate, consider the characteristic polynomial in the previous example again:

<math display="block">\det\!\begin{pmatrix}\lambda-1&-2\\-3&\lambda-4\end{pmatrix}.</math>

If one substitutes the entire matrix {{mvar|A}} for {{mvar|λ}} in those positions, one obtains

<math display="block">\det\!\begin{pmatrix} \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix} - 1 & -2 \\ -3 &\begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix} - 4\end{pmatrix},</math>

in which the "matrix" expression is simply not a valid one. Note, however, that if scalar multiples of identity matrices
instead of scalars are subtracted in the above, i.e. if the substitution is performed as

<math display="block"> \det\!\begin{pmatrix} \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix} - I_2 & -2I_2 \\ -3I_2 & \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix} - 4I_2 \end{pmatrix},</math>

then the determinant is indeed zero, but the expanded matrix in question does not evaluate to <math>A I_n-A</math>; nor can its determinant (a scalar) be compared to ''p''(''A'') (a matrix). So the argument that <math>p(A) = \det(A I_n - A) = 0</math> still does not apply.

Actually, if such an argument holds, it should also hold when other [[multilinear form]]s instead of determinant is used. For instance, if we consider the [[Permanent (mathematics)|permanent]] function and define <math>q(\lambda) = \operatorname{perm}(\lambda I_n - A)</math>, then by the same argument, we should be able to "prove" that {{math|1=''q''(''A'') = 0}}. But this statement is demonstrably wrong: in the 2-dimensional case, for instance, the permanent of a matrix is given by

<math display="block">\operatorname{perm}\!\begin{pmatrix} a & b \\ c & d \end{pmatrix} = ad + bc.</math>

So, for the matrix {{mvar|A}} in the previous example,

<math display="block">\begin{align}
q(\lambda) & = \operatorname{perm}(\lambda I_2 - A) = \operatorname{perm}\!\begin{pmatrix} \lambda - 1 & -2 \\ -3 & \lambda-4 \end{pmatrix} \\[6pt]
& = (\lambda - 1)(\lambda - 4) + (-2)(-3) = \lambda^2 - 5\lambda + 10.
\end{align}</math>

Yet one can verify that

<math display="block">q(A) = A^2 - 5A + 10I_2 = 12I_2 \neq 0.</math>

One of the proofs for Cayley–Hamilton theorem above bears some similarity to the argument that <math>p(A) = \det(AI_n-A) = 0</math>. By introducing a matrix with non-numeric coefficients, one can actually let {{mvar|A}} live inside a matrix entry, but then <math>A I_n</math> is not equal to {{mvar|A}}, and the conclusion is reached differently.