$L_H(X)$ is real vector space, Please help demonstrate that applies: $L_H(X)$ is real vector space, where $X$ is Hilbert space (real or complex), and $L_H(X)$ the set of all hermitian operator on $L(X).$
Thanks for your help and your attention. Thanky very much
 A: If $A, B \in L_H(X)$, we have $A^\dagger = A$ and $B^\dagger = B$; then for $a, b \in \Bbb R$, we need to show that $aA + bB \in L_H(X)$; that is, that $(aA + bB)^\dagger = aA + bB$.  This is most directly done by falling back on the fundamental definitions:  we recall that a complex Hilbert space such as $X$ is equipped with an inner product $\langle \cdot, \cdot \rangle$ which maps $X \times X \to \Bbb C$ and obeys, for all $x_1, x_2, y_1, y_2 \in X$ and $\alpha \in \Bbb C$:
$\langle x_1, y_1\rangle = \overline{\langle y_1, x_1 \rangle}, \tag{1}$
where $\bar z$ denotes the complex conjugate of any $z \in \Bbb C$;
$\langle x_1, y_1 + y_2 \rangle = \langle x_1, y_1 \rangle + \langle x_1, y_2 \rangle, \tag{2}$
$\langle x_1+ x_2, y_1\rangle = \langle x_1, y_1 \rangle + \langle x_2, y_1\rangle, \tag{3}$
$\langle x_1, \alpha y_1 \rangle = \alpha \langle x_1, y_1 \rangle, \tag{4}$
$\langle \alpha x_1,  y_1 \rangle = \bar \alpha \langle x_1, y_1 \rangle, \tag{5}$
$\langle x_1, x_1 \rangle \ge 0, \tag{6}$
$\langle x_1, x_1 \rangle = 0 \Leftrightarrow x_1 = 0. \tag{7}$
(1)-(7) define a norm $\Vert \cdot \Vert$ on $H$ via the equation
$\Vert x_1 \Vert^2 = \langle x_1, x_1 \rangle, \tag{8}$
where we take $\Vert x_1 \Vert \ge 0$ in (8); $\Vert \cdot \Vert$ defines the topology on $H$, and since $X$ is Hilbert, it is complete with respect to this norm (i.e., Cauchy sequences defined with respect to $\Vert \cdot \Vert$ converge to limits in $X$).  These are basic facts concerning the inner product $\langle \cdot, \cdot \rangle$ on $X$; I re-iterate them here for the sake of completeness; furthermore, denoting by $L(X)$ the set of all bounded operators on $X$, the Hermitian adjoint $C^\dagger$ of any $C \in L(X)$ is best defined via the inner product, though in the event that $\dim X < \infty$, so that elements of may be thought of as matrices of finite size, we can simply take $C^\dagger$ to be the conjugate transpose $\bar C^T$ of $C$ provided $C$ is expressed in terms of an orthonormal basis; but when $\dim X = \infty$, it is better to work directly from $\langle \cdot, \cdot \rangle$, thus:  $C^\dagger \in L(X)$ is defined to be the unique operator such that
$\langle C^\dagger x_1, y_1 \rangle = \langle x_1, Cy_1 \rangle \tag{9}$
for any all $x_1, y_1 \in X$.  (That $C^\dagger$ may be uniquely so defined is a standard fact whose validation I leave to my reader's research.)  It is easy to develop properties of $C^\dagger$ from (9) in a coordinate-independent, basis-free way, viz., $C^{\dagger \dagger} = C$, since we have 
$\langle x_1, Cy_1 \rangle = \langle C^\dagger x_1, y_1 \rangle = \overline{\langle y_1, C^\dagger x_1 \rangle} = \overline{\langle C^{\dagger \dagger} y_1, x_1 \rangle} = \langle x_1, C^{\dagger \dagger} y_1\rangle \tag{10}$
holding for all $x_1, y_1 \in X$.  (10) is indicative of the kind of argument we need to show that $aA + bB \in L_H(X)$; indeed we have
$\langle (aA + bB)^\dagger x_1, y_1 \rangle = \langle x_1, (aA + bB)y_1 \rangle = \langle x_1, aA y_1 \rangle + \langle x_1, bBy_1 \rangle$
$= a \langle x_1, Ay_1 \rangle + b\langle x_1, By_1 \rangle = a\langle A^\dagger x_1, y_1 \rangle + b\langle B^\dagger x_1, y_1 \rangle = \langle (aA^\dagger + bB^\dagger)x_1, y_1 \rangle, \tag{11}$
where we have used (9), (4), (5) and the fact that $a, b \in \Bbb R$, so that $a = \bar a$, $b = \bar b$, in (11).  Since (11) holds for all $y_1 \in X$, we must have
$(aA + bB)^\dagger x_1 = (aA^\dagger + bB^\dagger)x_1 \tag{12}$
for all $x_1 \in X$, whence
$(aA + bB)^\dagger = aA^\dagger + bB^\dagger = aA + bB, \tag{13}$
since $A^\dagger = A$ and $B^\dagger = B$; this proves the desired result, i.e., that $L_H(X)$ is a real linear subspace of $L(X)$.  QED.
Hope this helps.  Cheers,
and as always,
Fiat Lux!!!
