$A$ is a symmetric real matrix. Show that there is $B$ such that $B^3=A$ I'm having trouble with this question, I'd like someone to point me in the right direction.
let $A$ be a n by n matrix with real values. 
show that there is another n by n real matrix $B$ such that $B^3=A$, and that $B$ is symmetric. Are there more matrices like this $B$ or is it the only one?
What I was thinking:
I don't have a clear way to solve it. I think we need to use the fact that if a real matrix is symmetric, then it is normal, and so has an orthonormal basis of eigenvectors...Other then that I don't really know anything.
 A: Note that, every symmetric $A\in\mathbb R^{n\times n}$ matrix is diagonalisable, it has real eigenvalues $d_1,\ldots,d_n$, and its diagonalization is realised with an orthogonal matrix $U$, i.e.,
$$
A=U^TDU, 
$$ 
where $D=\mathrm{diag}(d_1,\ldots,d_n)$, and $U^TU=I$. Now let
$$
B=U^T\mathrm{diag}(d_1^{1/3},\ldots,d_n^{1/3})U.
$$
Clearly $B^3=A$ and
$$
B^T=\big(U^T\mathrm{diag}(d_1^{1/3},\ldots,d_n^{1/3})U\big)^T=U^T\mathrm{diag}(d_1^{1/3},\ldots,d_n^{1/3})U=B.
$$
Hence $B$ is symmetric.
A: Note that $\forall \mathbf{M}\in\mathbb{R}^{n\times n}$ such that $\mathbf{M}$ is symmetric, we have $\mathbf{M}=\mathbf{P}\mathbf{\Lambda}\mathbf{P}^{-1}=\mathbf{P}\mathbf{\Lambda}\mathbf{P}^{T}$, for some orthogonal matrix $\mathbf{P}$.
Therefore we have $\mathbf{B}^{3}=\left(\mathbf{P}\mathbf{\Lambda}_{\mathbf{B}}\mathbf{P}^{T}\right)^{3}=\mathbf{P}\mathbf{\Lambda}_{\mathbf{B}}^{3}\mathbf{P}^{T}$, and $\mathbf{A}=\mathbf{P}\mathbf{\Lambda}_{\mathbf{A}}\mathbf{P}^{T}$, therefore we have $\mathbf{\Lambda}_{B}^{3}=\mathbf{\Lambda}_{\mathbf{A}}$, where $\mathbf{\Lambda}_{\mathbf{A}}=\operatorname{diag}(\lambda_{1},\dots,\lambda_{n})$ and $\lambda_{i}$ are the eigenvalues of $\mathbf{A}$.
Therefore we have:
$$\mathbf{B}=\mathbf{P}\operatorname{diag}(\sqrt[3]{\lambda_{1}},\dots,\sqrt[3]{\lambda_{n}})\mathbf{P}^{T}$$
A: I would first like to thank Mariano Suarez-Alvarez in advance for pointing me in the right direction.
if $A$ is symmetric over $\mathbb R$, then it is diagonalizable:
$A=PDP^{-1}$ such that $D$ is diagonal.
let $B = PD_2P^{-1}$ such that $D_2$ is a diagonal matrix whos values are the third root of the matrix $D$.
so we get $B^3 = PD_2^{3}P^{-1} = PDP^{-1}=A$
A: If $B$ satisfies $B^3=A$, then $\{ B, \alpha B, \overline{\alpha} B \}$ are solutions, where $\alpha = \exp \left(\frac{2\pi i}{3}\right)$ and where $\overline{\alpha} = \exp \left(-\frac{2\pi i}{3}\right)$ is the conjugate of $\alpha.$
