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If $A$ is a real Symmetric Matrix of order $n (\geq 2)$ , then there exists a symmetric Matrix $B$ such that $B^{2k+1} = A$.

Is the statement true? I think the statement is true.

My Attempt : I have got an idea to prove it. As $A$ is real symmetric Matrix it can be orthogonally diagonalizable.So we can write $A = PDP^T$ Where the $D$ is a real diagonal matrix whose diagonal elements are the eigen values of $A$. we can also write $A = PD'P^T . PD'P^T...….PD'P^T(2k+1$ times). Where the $D'$ is a diagonal matrix with the $ii$ th element is $a_{ii}^{1/ 2k+1}$ if the $ii$th element of $D$ is greater than $0$ and the $ii$ th element will be $-(|a_{ii}|)^{1/ 2k+1}$ if the $ii$th element of $D$ is less than $0$.

Have I gone correct? Can anyone please tell me If there is any mistake?

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For every real number $a$ there is a real number $b$ such that $b^{2k+1}=a$. Apply this to each of the diagonal elements of $D$ to get you $D'$.

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  • $\begingroup$ Yea..That's what I applied. Can you please check if my attempt is correct or not? @Kavi Rama Murthy $\endgroup$ – cmi Jan 16 at 6:26
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    $\begingroup$ @cmi from what I see you have not defined $D'$ properly. In fact $k$ doesn't seem to appear in your construction of $D'$. That is the reason I posted my answer. $\endgroup$ – Kavi Rama Murthy Jan 16 at 6:35
  • $\begingroup$ I edited...........@Kavi Rama Murthy $\endgroup$ – cmi Jan 16 at 6:53

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