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

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    $\begingroup$ Because $x \mapsto \sqrt[n]{x}$ is bijective. And $T$ can be diagonalised... $\endgroup$ – copper.hat Mar 18 at 20:26
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    $\begingroup$ If all you have is a real scalar product you don't have a notion of hermitian. You mean $T$ is symmetric? The latter means that $T$ can be diagonalized by a orthogonal transformation and the eigenvalues are real. Then you need to find the $n$-th root of said numbers. If $n$ is odd there's always a real solution. $\endgroup$ – lcv Mar 18 at 20:28

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

$O:V \to V, \tag 2$

$OO^t = O^tO = I, \tag 3$

$OTO^t = D = \text{diag} (t_1, t_2, \ldots, t_m), \; t_i \in \Bbb R, \; 1 \le i \le m, \tag 4$


$m = \dim_{\Bbb R}V; \tag 5$

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

$R = \text{diag} (\rho_1, \rho_2, \ldots, \rho_m); \tag 7$


$R^n = \text{diag} (\rho_1^n, \rho_2^n, \ldots, \rho_m^n) = \text{diag} (t_1, t_2, \ldots, t_m) = D; \tag 8$

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

$S = O^tRO. \tag{11}$


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