Real Inner Product Space, Hermitian Operator $T = S^{n}$ for n odd Let V be a finite dimensional inner product space over $\mathbb{R}$, and $T: V\rightarrow V$ hermitian.
Suppose n is an odd positive integer.
Want to show:
$\exists S:V\rightarrow V $ such that $T = S^{n}$
Here, I know that T is Hermitian if $T = T^{*}$ its complex conjugate and so, $T = T^{t}$ because it is a real operator.
I am having trouble with this question because I don't know where the criteria that n is odd will come in. Will the reasoning be similar to this? https://math.stackexchange.com/a/89715/651806
 A: On a finite-dimensional real inner product space, the notions of hermitian and symmetric for operators coincide; that is,
$T^\dagger = T^t = T; \tag 1$
since $T$ is symmetric, it may be diagonalized by some orthogonal operator
$O:V \to V, \tag 2$
$OO^t = O^tO = I, \tag 3$
$OTO^t = D = \text{diag} (t_1, t_2, \ldots, t_m), \; t_i \in \Bbb R, \; 1 \le i \le m, \tag 4$
where 
$m = \dim_{\Bbb R}V; \tag 5$
since each $t_i \in \Bbb R$, for odd $n \in \Bbb N$ there exists 
$\rho_i \in \Bbb R, \; \rho_i^n =t_i; \tag 6$
we observe that this assertion fails for even $n$, since negative reals do not have even roots; we set
$R = \text{diag} (\rho_1, \rho_2, \ldots, \rho_m); \tag 7$
then
$R^n =  \text{diag} (\rho_1^n, \rho_2^n, \ldots, \rho_m^n) = 
\text{diag} (t_1, t_2, \ldots, t_m) = D; \tag 8$
it follows that
$T = O^tDO = O^tR^nO = (O^tRO)^n, \tag 9$
where we have used the general property that matrix conjugation preserves products:
$O^tAOO^TBO = O^tABO, \tag{10}$
in affirming (9).  We close by simply setting
$S = O^tRO. \tag{11}$
