Let V be an inner product space and $(v_1, \ldots, v_n)$ and $(w_1, \ldots, w_n)$ be orthonormal bases of $V$. Let $F \in$ End$(V)$ be the distinct mapping with $F(v_j) = w_j$ for alle $j \in \{1, \ldots, n\}$. Prove that $F$ is orthogonal.
My Proof:
We will show that $F$ is a bijection and preserves the inner product. Because the bases are orthonormal and therefore orthogonal, we obtain $$ \langle F(v_i), F(v_j) \rangle = \langle w_i, w_j \rangle = 0 = \langle v_j, v_j \rangle $$ for all $i, j \in \{1, \ldots, n\}$ with $i \not= j$.
To show that $F$ is a bijection it suffices to show its injectivity, since in finite spaces, injectivity, surjectivity and bijectivity are equivalent. For all $v \in V$ we obtain $$ F(v) = 0 \implies \| F(v) \| = 0 \implies \| v \| = 0 \implies v = 0 \implies \ker(F) = \{0\}. \square $$
My Question:
My proof was marked to be correct, but why does it suffice to do this procedure for the bases, and doesn't have to be done for a arbitrary vector in $V$?