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 be an matrix with characteristic equation . Then .

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

is diagonalisable, so there exists a basis of eigenvectors . It will suffice to show that the linear map corresponding to sends everything to zero.

Any vector can be written , so by the linearity of it suffices to show that each component is mapped to the zero vector. But clearly , where is the eigenvalue of .

So we are done. 🙂

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January 6, 2009 at 4:17 pm

Justin DrakeCayley-Hamilton? Cauchy-Davenport is some sort of combinatorial sum-sets thing isn’t it?

January 16, 2009 at 8:40 pm

tloveringIndeed you’re correct. The error has now been fixed. Thanks.

January 20, 2009 at 7:24 pm

compactorangeActually, at this point, we can go further. When the coefficient ring is a (algebraically closed) field, we can finish the proof(s) of the CH by introducing the Zariski topology.