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Let $B \in M_n$ be any matrix with minimal polynomial $m(x)=(x+1)^2$. Find $x,y \in \mathbb C$ such that $$B^{150}=xB+yI$$

I don't know how to approach this problem, I have an idea that one should use the Lagrange-Hermite Interpolation here but I don't get the idea on how one should apply it so maybe there's other method. Any help and hints would be much much appreaciated. Thanks!

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2 Answers 2

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You know that $(B+I)^2=0$, thus $(B+I)^n = 0$ for $n \ge 2$. Write $B^{150} = ((B+I)-I)^{150}$, expand using the binomial formula and only keep the terms where $(B+I)$ is raised to a power lower than $2$.

If my computation is correct, you will find $B^{150} = -149I - 150B$

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Hint: Write $B^{n}=x_nB+y_nI$. Use $B^2=-2B-I$ to find recurrences for $x_n$ and $y_n$.

Alternatively, write $T^n = q_n(T)(T+1)^2+x_n T+y_n$. Evaluate this and its derivative at $T=-1$ to find $x_n$ and $y_n$.

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