Cayley Hamilton Theorem states that if $A$ is an $n \times n$ matrix over the field $F$ then $p(A) = 0$.
We note that $p(\lambda) = \det(\lambda I – A)$. Hence, why can't we just substitute $\lambda$ with $A$ and directly prove Cayley-Hamilton Theorem by saying that $p(A) = p(\lambda) = \det(\lambda I – A) = \det(AI – A) = 0$?
Best Answer
There is another way to see that the proof must be flawed: by finding the interesting consequences this proof technique has. If the proof would be valid, then we would also have the following generalisation:
This has the following amazing consequence:
In particular, this proves that we have $$ \pmatrix{1 & 0 \\ 0 & 0}^2 = \pmatrix{0 & 0 \\ 0 & 0}. $$ This comes to show just how wrong the proof is!