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When do polynomials maintain similarity? Does this result from Cayley–Hamilton theorem?

Similar matrices have the same eigenvalues, but for instance $A$ and $A-\lambda I$ don't have the same eigenvalues. The later has $A$'s eigenvalues shifted by $\lambda$.

I'm missing something here.

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Similarity is respected by addition and raise to power, so it is preserved by any polynomial. –  anonymous Nov 7 '12 at 11:03
But does that follow from the Cayley–Hamilton theorem? –  Robert S. Barnes Nov 7 '12 at 11:24
I'm not sure what you mean by polynomials maintain similarity, but in any case Cayley–Hamilton is a far deeper fact and has nothing to do with this. –  Marc van Leeuwen Nov 7 '12 at 11:37
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1 Answer

If you mean by "polynomials maintain similarity" that if you substitute two similar matrices into the same polynomial then the resulting polynomials will be similar, then this fact simply reflects the fact that substituting linear maps $V\to V$ into a polynomial is well defined.

If $A,A'$ are similar $n\times n$ matrices, then there is some linear map $f:V\to V$ for a vector space of dimension $n$ such that if has matrix $A$ in some basis $\mathcal B$ and matrix $A'$ in another basis $\mathcal B'$. Now if $P=\sum_{i=0}^dp_iX^i$ is some polynomial, then substitution of $f$ for $X$ into $P$ gives $\phi=\sum_{i=0}^dp_if^i$, where $f^i$ is the composition of $i$ copies of $f$ (by convention this gives the identity of $V$ when $i=0$). The matrix of $f^i$ in the basis $\mathcal B$ is the matrix power $A^i$, and its matrix basis $\mathcal B'$ is $(A')^i$. Therefore the matrix of $\phi$ in the basis $\mathcal B$ is $\sum_{i=0}^dp_iA^i$ and in the basis $\mathcal B'$ it is $\sum_{i=0}^dp_i(A')^i$; these two matrices are similar by definition.

The fact that $A-\lambda I$ is not similar to $A$ is simply because subtracting a multiple of the identity has nothing to do with a change of basis.

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