# 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

• Because $x \mapsto \sqrt[n]{x}$ is bijective. And $T$ can be diagonalised... – copper.hat Mar 18 at 20:26
• 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. – 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

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

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

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

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

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