Spin groups and Clifford Algebra

Question

Consider $\Bbb R^n$ with standard norm and its Clifford algebra $Cl(\Bbb R^n)$. It decomposes as $Cl(\Bbb R^n)=Cl_0(\Bbb R^n)\oplus Cl_1(\Bbb R^n)$ (even part and odd part). The group $\text{Spin}(n) \subset Cl_0(\Bbb R^n)$ is defined to be $\{v_1 \cdots v_r : v_i\in \Bbb R^n, |v_i|=1, r \textrm{ even}\}$ (the multiplication $v_1\cdots v_r$ is taken in $Cl(\Bbb R^n)$). (I am following Lawson, Michelson's book Spin Geometry.) It is known that $Cl_0(\Bbb R^n)$ is isomorphic to $Cl(\Bbb R^{n-1})$, so we can regard $\text{Spin}(n)\subset Cl(\Bbb R^{n-1})$. Theorem 8.1 (p.50) of the book asserts that $\text{Spin}(3)\cong SU(2)$, $\text{Spin}(4)\cong SU(2)\times SU(2)$, etc. In the proof, it is written that any representation of $Cl(\Bbb R^{n-1})$ on $\Bbb C^m$ or $\Bbb H^m$ can be assumed to have the property that, when restricted to the group $\text{Spin}(n)$, it preserves the standard hermitian inner product. But I can't see why. Why is this assumption possible?