In fact ,there are many question about production of reflection in mathematics . But seemly, they are talking it in plane.
As I know ,if V is a n-dim real vector space, the elements of O(V) is reflection and rotation of V. But I don't know if use matrix to present element of O(V) , how to distinguish what are reflection ,what are rotation ? I mean what matrix is rotation ,what is reflection. Then, how to show the product of two reflections is rotation from the normal vector of hyperplane of reflection to another.
If you're working with a real inner product on $V$ with positive definite signature then we can describe the product of two reflections geometrically.
First, some linear algebra. Say $V$ is an internal direct sum of subspaces, $V=V_1\oplus V_2$. Suppose further that $A$ is a linear map on $V$ which preserves $V_1$ and $V_2$. That is, it restricts to linear maps $A_1$ and $A_2$ on them. Then we may use the notation $A=A_1\oplus A_2$. (Conversely, given linear maps $A_1$ and $A_2$ on $V_1$ and $V_2$, there is a unique linear map $A$ on $V$ that restricts to them.) If we can do the same thing for another linear map $B$ on $V$, then we have $A+B=(A_1+B_1)\oplus (A_2+B_2)$ and $AB=(A_1B_1)\oplus(A_2B_2)$. In other words, there is an algebra homomorphism
$$ \mathrm{End}(V_1)\times\mathrm{End}(V_2)\to \mathrm{End}(V_1\oplus V_2). $$
Thus, to compose $A$ and $B$ we may find subspaces they stabilize and compose the two transformations "componentwise." This works with more than two components too.
Suppose $A$ is a reflection on $V$ across a hyperplane $\Pi$ perpendicular to some one-dimensional subspace $P$. Then $A$ stabilizes $P$ and $\Pi$. Moreoever, it acts as the identity on $\Pi$ and the negation map on $P$. Then suppose $B$ is another reflection, this time across a hyperplane perpendicular to a different one-dimensional subspace $L$. Now let's write $V=P\oplus L\oplus W$ where $W$ is the orthogonal complement of $\mathrm{span}(P,L)=P+L$. Then $A$ and $B$ act as $\mathrm{id}$ on $W$, so $AB$ acts as $\mathrm{id}$ on $W$ as well, and it suffices to see how $AB$ acts on the two-dimensional subspace $P+L$.
For this, we can draw a picture:
This plane represents the span of $P$ and $L$. The axis $P'$ and $L'$ are the lines perpendicualr to $P$ and $L$ respectively. Starting with a vector $x$, flip it across $P'$ to get $y$, then flip $y$ across $L'$ to get $z$. The effect of going from $x$ to $z$ is moving by an angle of $2(a+b)$. But that's exactly twice the angle between $P'$ and $L'$, which is independent of the original vector $x$, so all vectors are rotated by an angle of $2(a+b)$.
The orthogonal group $O(V)$ has two connected components. The connected component of the identity is $SO(V)$, the group of rotations. The other connected component includes hyperplane reflections, but it also generally includes other transformations which are not hyperplane reflections (such as $-I_3\in O(3)$, which has no nonzero fixed points, let alone a hyperplane's worth). In any case, these two connected components are precisely the fibers of $\det:O(V)\to\{\pm1\}$. That is, an element of $O(V)$ is a rotation if it has determinant $1$ and has determinant $-1$ otherwise.