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My Classmate asked me two following question this morning. But I still can't figure out proper solutions...

(1) If $A \in R^{nxn}$, satisfy $A^{3}+A+I_n=0$, find$a,b,c\in R$ such that$(A^{2}-A-I_n)^{-1}=aI_n+bA+cA^{2}$

My Approach

By Cayley Hamilton Theorem, $$0=A^{3}+A+I_n=(A^{2}-A-I_n)(A+I_n)+3A+2I_n$$ then I don't know how to reorder the equation to fit the $(A^{2}-A-I_n)^{-1}$ approach.

(2) If $A\in R^{nxn}$and$(A-I_n)(A-9I_n)(A-16I_n)=0$, find a matrix $B=aI_n+bA+cA^2(a,b,c\in R)$ such that $B^{2}=A$

My Approach

Either I try to solve it by (1) brute-force or (2) minimal polynomial, I have no idea how to get the proper solution. The only thing I know is that minimal polynomial is only satisfied while A is diagonalizable. Is there any proper solution? or which step should start with?

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up vote 1 down vote accepted

For (2), $A$ has eigenvalues $1$, $9$ and $16$. Square roots of these are $1$, $3$ and $4$. Take $B = p(A)$ where $p(1) = 1$, $p(9) = 3$ and $p(16) = 4$, and it will do. Lagrange interpolation gives $$p(x) = \dfrac{(x-9)(x-16)}{(1-9)(1-16)} + 3 \dfrac{(x-1)(x-16)}{(9-1)(9-16)} + 4 \dfrac{(x-1)(x-9)}{(16-1)(16-9)}$$

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Thanks! So it is not related to minimal polynomial solution, is it? – lucasKoFromTW Jan 12 '13 at 15:44
The eigenvalues come from the minimal polynomial, and the fact that the factors in the minimal polynomial are not repeated is also used. – Robert Israel Jan 13 '13 at 4:05

Here is a brute-force approach, spelling out $\left(A^2-A -I_n\right) \cdot \left(A^2-A -I_n\right)^{-1} = I_n$: $$ \begin{eqnarray} (A^2 - A - I_n) (a I_n + b A + c A^2) &=& \left( (b - c)I_n + c A \right)\left(A^3+A+I_n\right) + \\&& \left(c-a-b\right) I_n - \left(a + 2 b \right)A + \left(a-b-2c\right)A^2 \end{eqnarray} $$ Thus we have the following linear system: $$ c - a - b = 1 \quad a+2b=0 \quad a - b - 2c = 0 $$with the solution $a=4$, $b=-2$ and $c=3$.

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Thanks a lot! But how come the condition A^3+A+I_n = 0 will not affect the result gained by the brute force? And why can't I use brute force to solve the problem 2? – lucasKoFromTW Jan 11 '13 at 16:52

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