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a) Let $A$ is of nilpotent of degree $k$ then according to the condition ${A}^{k}={B}^{2k}=0$ But I can't say anything about existence of such $B$

I have no idea on $b$, $c$, please help.

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    Hint for a: $B$ would be nilpotent. Eigenvalues? Characteristic polynomial? Hint for b,c: Diagonalizability theorems: try to understand the case when $A$ is diagonal first.2012-07-21

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(a) Can you show that if $\,A\,$ is a $\,n\times n\,$ nilpotent matrix , say$\,A^k=0\,$, then $\,k\leq n\,$? This answers this section as $\,B\,$ is nilpotent, but then $\,B^2=0\,$ , so if $\,A\neq 0\,$ then the claim isn't true.

(b) Since $\,A\,$ is symmetric positive definite his eigenvalues are positive real, so we can write (wrt some basis of eigenvectors) $A=\begin{pmatrix}a_1&0\\....&....\\0&a_n\end{pmatrix}\,\,,\,a_i>0$ Well, finding $\,B\,$ now is easy...

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  1. $B$ would be nilpotent with an index $\geq 3$ (since $B\neq 0$, $B^2\neq 0$).
  2. Use the fact that $A$ is diagonalizable: $A=P^tDP$ with $P$ orthogonal and $D$ diagonal. The elements of $D$ are positive, and take $B=P^t\operatorname{diag}(\sqrt{\lambda_j})P$.
  3. Take $B=P^t\operatorname{diag}(\sqrt[3]{\lambda_j})P$ (the eigenvalues are real).