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Let $M(n,\mathbb R)$ denote set of all $n\times n$ matrices over $\mathbb R$.Which are true:

1.If $A\in M(2,\mathbb R)$ is nilpotent and non-zero ,then there exists a matrix $B\in M(2,\mathbb R)$ such that $B^2=A$

2.If $A\in M(n,\mathbb R)$ is symmetric and positive definite ,then there exists a symmetric matrix $B\in M(n,\mathbb R)$ such that $B^2=A$

3.If $A\in M(n,\mathbb R)$ is symmetric ,then there exists a symmetric matrix $B\in M(n,\mathbb R)$ such that $B^3=A$

I dont know how to approach these?

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Here are some hints to get you started.

  1. If $A$ is nilpotent and $B^2 = A$, what can you say about higher powers of $B$? And which power of $A$ is guaranteed to give you the zero matrix?

  2. Think about diagonalizing $A$. Can you compute a square root of a diagonal matrix?

  3. Think about diagonalizing $A$. Can you compute a cube root of a diagonal matrix?

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  • $\begingroup$ what is meant by cube root or square root of a matrix?can u please elaborate $\endgroup$ – Learnmore Dec 7 '14 at 4:55
  • $\begingroup$ for the first part $A^2=0$ $\endgroup$ – Learnmore Dec 7 '14 at 4:56
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    $\begingroup$ $B$ is a square root of $A$ if $B^2 = A$. $\endgroup$ – Hans Engler Dec 7 '14 at 14:02
  • $\begingroup$ I managed to come up with a counter-example for part (a) but couldn't proceed in the other two parts. Would you kindly post a complete solution? $\endgroup$ – Brozovic Jan 10 at 21:12
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  1. If A is $\begin{bmatrix}{0}&{1}\\{0}&{0}\end{bmatrix}$ clearly ${A}{A}=0$ and you can prove that is a counterexample.
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