How would one prove the Cayley-Hamilton theorem for a case when A=SDS^−1, where A is diagonal? I have absolutely no idea.
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$\chi_A(t)=\det(tA-I)=\det(tSDS^{-1}-I)=\det(S(tD-I)S^{-1})=\det(tD-I)=\chi_D(t)$.
Therefore, $\chi_A(A)=\chi_D(SDS^{-1})=S\chi_D(D)S^{-1}=0.$
That is, you just need to prove the Cayley–Hamilton theorem for diagonal matrices, which is easy.
Indeed, $\chi_D(t)=(t-d_{11})\cdots(t-d_{nn})$. Since $De_i=d_{ii}e_i$, we have $\chi_D(D)e_i=0$ for all $i$ and so $\chi_D(D)=0$.
lhf
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So for proving the Cayley-Hamilton theorem for diagonal matrices I am still slightly confused. I am not entirely sure about the Cayley-Hamilton proof as a whole. I know the concept that when we plug in the matrix to the characteristic equation it equals zero, and I have seen it work. But how do I prove it? – readyforthebettys Dec 02 '16 at 00:09
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Can I ask how you did the deconstruction of det(tA-I)? Particularly, how does it go to Det(S(tDS-I)S)=det(tD-I)? I think I now understand the proof of Cayley Hamilton with a diagonal matrix. – readyforthebettys Dec 04 '16 at 01:56
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@readyforthebettys, $\det (AB) = \det(A)\det(B)$, $SS^*=I$. – lhf Dec 04 '16 at 13:13
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Not sure what you mean. Take $$A=\begin{bmatrix}1&0\\0&2\end{bmatrix},\ \ \ S=\begin{bmatrix}1&0\\0&0\end{bmatrix}.$$ Then $p_A(t)=t^2-3t+2$. Also, $$ SAS^*=S=\begin{bmatrix}1&0\\0&0\end{bmatrix}, $$ so $$p_A(SAS^*)=\begin{bmatrix}0&0\\0&2\end{bmatrix}.$$
Martin Argerami
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I meant inverse, not *, sorry my bad! Misread the original question I was looking at / perplexed about – readyforthebettys Dec 01 '16 at 23:51