Attempting to solve a question from homework, relating to Jordan normal form:
Let $a \in \mathbb{C}, N \in M_n^\mathbb{C}$. $N$ is nilpotent of index 3. Prove that the matrix $S \in M_n^\mathbb{C}$, defined as follows: $$S=a\left(I_n + {1\over2}N - {1\over8}N^2\right)$$ satisfies: $$S^2 = a^2\left(I_n + N\right)$$
This part was trivial - and have solved it. The next part is as follows, which I am having trouble with:
Let $\lambda\in\mathbb{C}, \lambda\ne0$. Find a matrix $S \in M_3^\mathbb{C}$ such that $S^2=\lambda^2I_n+J_3$.
This was obviously trivial if the requirement was $\lambda^2(I_n+J_3)$, as this is simply a use of the results of the first part.
I've therefore tried as follows. We require a matrix $A$ such that $A^2=\lambda^2I_n+J_3$. $S$ is the matrix that satisfies the first part.
I'll try to find $A$ such that $A^2 = S^2 - (\lambda^2-1)J_3$. I've tried completing the square, and have ended up with:
$$\left(S-{\lambda^2-1\over2}J_3\right)^2-\left(\left(\lambda^2-1\over2\right)J_3\right)^2$$
Which would've been conventient had the nilpotency index of N in the first part been 2.
Would appreciate any assistance / hints. Thanks.
I suppose that $J_3$ means the $3\times3$ Jordan block for the zero eigenvalue (correct me if I am wrong), so that $J_3^3=0$. You should try a matrix $S$ of the form $S = aI + bJ_3 + cJ_3^2$.