TITLE: Error in argument regarding the Cayley Hamilton theorem
QUESTION [7 upvotes]: I cannot spot the mistake in the following argument regarding the Cayley Hamilton theorem:
Let $A\in M_n$, then,
$$\begin{align*}
P_A(t)&=\det(tI-A)\\
&\implies P_A(A)=\det(AI-A)\\
&\implies P_A(A)=\det(0)\\
&\implies P_A(A)=0
\end{align*}$$

REPLY [8 votes]: 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 $\mathbf M_n(K)$."