In Part IA Vectors and Matrices this theorem was proved, but the following proof (which I thought of during the lecture and seemed most natural) wasn’t lectured, or if it was it wasn’t clear to me that this was the approach taken.

Claim: Let $A$ be an $n\times n$ matrix with characteristic equation $f(x)=0$. Then $f(A)=0$.

Proof: We prove it in the special case where $A$ is diagonalisable.

$A$ is diagonalisable, so there exists a basis of eigenvectors $e_1,e_2,...,e_n$. It will suffice to show that the linear map corresponding to $L= f(A)$ sends everything to zero.

Any vector can be written $v=v_1e_1+v_2e_2+....+v_ne_n$, so by the linearity of $L$ it suffices to show that each component is mapped to the zero vector. But clearly $L(v_je_j) = v_jL(e_j) = v_j f(\lambda_j)e_j = 0e_j = 0$, where $\lambda_j$ is the eigenvalue of $e_j$.

So we are done. 🙂