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. 🙂