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Given the characteristic polynomial and minimal polynomial of $A$ being $(x+2)^{6}(x-1)^{3}$ and $(x+2)^{2}(x-1)^{3}$, respectively. How do I determine the characteristic polynomial and minimal polynomial of $A-2A^{-1}$. I tried to algebraically rearrange this in order to have two separate determinant calculations but haven't had success so far. Also I have the Jordan Canonical Form of $A$ and must determine that of the $A-2A^{-1}$ matrix.

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  • $\begingroup$ would $A-2A^{-1}$ = $A-2I/A$=$A^{2}-2I$ be correct? $\endgroup$ – B Best Apr 1 '17 at 2:00
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From the characteristic polynomial you get that the $9$-dimensional space $A$ acts on decomposes as direct sum of generalised eigenspaces for $\lambda=-2$ (which has dimension$~6$) and for $\lambda=1$ (which has dimension$~3$). Since $0$ is not an eigenvalue, $A^{-1}$ and hence $A-2 A^{-1}$ are well defined, and the mentioned summands are stable for$~A-2 A^{-1}$ too. On these subspaces (the restriction of) $A-2 A^{-1}$ has unique eigenvalue $-2-2\times(-2)^{-1}=-2+1=-1$ for the former, respectively $1-2\times1^{-1}=1-2=-1$ for the latter. Since this is the same eigenvalue for $A-2 A^{-1}$, the whole space forms a single generalised eigenspace for it. These considerations already give the characteristic polynomial $(X+1)^9$ of$~A-2 A^{-1}$.

For the minimal polynomial, one may consider the largest Jordan block for a given eigenvalue$~\lambda$, whose size gives the exponent of $(X-\lambda)$ in the minimal polynomial. From the given minimal polynomial for$~A$ one gets $2$ as the size of the largest Jordan block for $\lambda=-2$, and $3$ the size of the unique hence) largest Jordan block for $\lambda=1$. All these blocks are also Jordan blocks of$~A-2 A^{-1}$ for its (unique) eigenvalue$~{-}1$, so the minimal polynomial of$~A-2 A^{-1}$ will be $(X+1)^3$. Alternatively, you may argue directly that $(A-2 A^{-1}+I)^3$ acts as $0$ on both generalised eigenspaces for$~A$, but $(\cdots)^2$ does not.

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$A$ is similar to a triangular matrix with a diagonal of $-2$ repeated $6$ times and $1$ repeated $3$ times. Therefore, $\det A= (-2)^6 \cdot 1^3 = 2^6$. For simplicity, let's assume $A$ is triangular.

We have $\det (A-2A^{-1}I) \det A = \det (A^2-2I)$.

The diagonal of $A^2$ is $(-2)^2$ repeated $6$ times and $1^2$ repeated $3$ times.

Therefore, the diagonal of $A^2-2I$ is $(-2)^2-2$ repeated $6$ times and $1-2$ repeated $3$ times and so $\det (A^2-2I)=-2^6$.

Therefore, $\det (A-2A^{-1}I)=-1$.

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  • $\begingroup$ So $det(xI-(A-2A^{-1}I))$ = $(x+1)$? (The characteristic polynomial) $\endgroup$ – B Best Apr 1 '17 at 2:13
  • $\begingroup$ @BBest, the characteristic polynomial has degree $9$ because $A$ is $9 \times 9$. $\endgroup$ – lhf Apr 1 '17 at 2:16
  • $\begingroup$ okay I just realized I screwed my question up asking for determinants rather than characteristic polynomials and minimal polynomials. But this definitely helped me. I just have to get the minimal polynomial now and use those to find the Jordan normal form of $A-2A^{-1}$ $\endgroup$ – B Best Apr 1 '17 at 2:21

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