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

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?

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 $$D'$$.
• @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. – Kavi Rama Murthy Jan 16 at 6:35