A question in my linear algebra textbook asks me to prove that $cI_2$ is not diagonalizable. Since an $n\times n$ matrix $A$ is diagonalizable only if it has $n$ linearly independent eigenvectors, we know that $cI_2$ is not diagonalizable. However, using a different approach, I reached the opposite conclusion and can't figure out where I went wrong.
By definition, a matrix $A$ is diagonalizable if $P^{-1}AP = D$, and $D$ is diagonal. So,
$\begin{align} D & = P^{-1}AP \\ & = P^{-1}(cI_2)P \\ & = c(P^{-1}IP)\\ & = c(P^{-1}P)\\ & = cI \end{align}$
Since $cI$ is diagonal, $D$ is diagonal, and $cI$ is therefore diagonalizable. A similar approach I thought of is to let $P=I$ so that
$\begin{align} D & = P^{-1}(cI_2)P \\ & = c(I^{-1}II)\\ & = cI^3\\ & = cI \end{align}$