Error in argument regarding the Cayley Hamilton theorem I cannot spot the mistake in the following argument regarding the Cayley Hamilton theorem:
Let $A\in M_n$, then,
$$\begin{align*}
P_A(t)&=\det(tI-A)\\
&\implies P_A(A)=\det(AI-A)\\
&\implies P_A(A)=\det(0)\\
&\implies P_A(A)=0
\end{align*}$$
 A: I'll just spell out Nick Strehlke's comment in the case $n=2$: 
$$A=\begin{pmatrix}a&b\\ c&d\end{pmatrix},$$ 
$$M(t):=tI-A=\begin{pmatrix}t-a&-b\\-c&t-d\end{pmatrix},$$ 
$$M(A)=\begin{pmatrix}A-aI&-bI\\-cI&A-dI\end{pmatrix}.$$ 
The Cayley-Hamilton states that the determinant of the above two by two matrix (whose coefficients are two by two matrices) is the $0$ two by two matrix (with scalar coefficients). 
More generally, if $A$ is in $\mathbf M_n(K)$, then $M(A)$ is in $\mathbf M_n(K[A])$, and its determinant is in $K[A]$. [The notation is, I hope, self-explanatory.] 
EDIT. Thank you very much to darij grinberg for having pointed out an error (which was especially unfortunate in an answer supposed to correct an error...). To let darij's comment remain understandable, here is the uncorrected version of the erroneous phrase: 
"... if $A$ is in $\mathbf M_n(K)$, then $M(A)$ is in $\mathbf M_n(\mathbf M_n(K))$, and its determinant is in $\mathbf M_n(K)$."
A: In the first step, you obtain the characteristic polynomial $P_A(t)$ which is obtained by calculating the determinant, det$(tI-A)$. And then, the next implication is the mistake that a multivariate polynomial obtained from the scalar entries of the matrix $A$ can  be written as det$AI-A$. The reason is simply because Cayley-Hamilton theorem uses two different polynomial types but same scalar coefficients. One is 
$$
P_A(t) = t^n+a_{n-1}t^{n-1}+\ldots+a_1t + a_0 = 0 \in\mathbb{R}
$$
The other one is 
$$
A^n+a_{n-1}A^{n-1}+\ldots+a_1A + a_0I = \mathbf{0} \in\mathbb{R}^{n\times n}
$$
Therefore, those are not compatible. 
I will stop before mention the word ring :)
