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 $V$?


1 Answer 1


It suffices to do this procedure for the bases because from it we have easily the result for arbitrary vector in $V$. In fact, let $x=\sum\limits_{i=1}^n x_i v_i$ and $y=\sum\limits_{i=1}^n y_i v_i$, so

$$\langle F(x),F(y)\rangle=\sum_{i=1}^n\sum_{j=1}^n x_iy_j\langle F(v_i),F(v_j)\rangle=\sum_{i=1}^n\sum_{j=1}^n x_iy_j\langle v_i,v_j\rangle=\langle x,y\rangle$$


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