Isometry on the spanning set is an isometry, intuition. I was reading and came across the following statement.

Let $C$ and $D$ be vector spaces equipped with bilinear forms, and $F: C \to D$ a linear map. Say $X_i$ is a spanning set of vectors for $C$ and $F(X_i)$ spans $D$. Suppose $F$ is "an isometry on the spanning set," i.e., $B(X_i, X_j) = B(F(X_i), F(X_j))$. Then $F$ is an isometry.

To me, this is not all that clear. Could someone provide me intuition for this statement, hopefully geometric, if possible?
 A: Intuitively: For simplicity assume the spanning set is in fact a basis. By the conditions, first of all the lengths $\sqrt{B(X_i,X_i)}$ of the basis vectors remain unchanged. After that, the invariance of $B(X_i,X_j)$ implies that all angles between the vectors remain unchanged. Thus the whole $F$ is "rigidly" constrained.
Admittedly, one could still imagine that some fliping takes place among the base vectors that does something harmful to some linear combination of them, but, hey, that's what you get from relying on intuition when calculatoin is a one-liner (if the line is long enough)
$$\begin{align}\left|F\left(\sum c_iX_i\right)\right|&=\sqrt{B\left(F\left(\sum c_iX_i\right),F\left(\sum c_iX_i\right)\right)}\\
&=\sqrt{B\left(\sum c_i F(X_i),\sum c_iF(X_i)\right)}\\
&=\sqrt{\sum_i\sum_j c_i\overline{c_j} B(F(X_i),F(X_j)) }\\
&=\sqrt{\sum_i\sum_j c_i\overline{c_j} B(X_i,X_j) }\\
&=\sqrt{B\left(\sum c_i X_i,\sum c_iX_i\right)}\\
&=\left|\sum c_iX_i\right|\end{align}$$
A: Your forms are bilinear, so intuitively, this is just saying that if you have a linear map $F:C \to D$, then preserving the bilinear form on the basis vectors is enough to guarantee that it preserves the bilinear form on the entire space.
This is just because of the linearity.  A linear map is completely determined by where it sends the basis vectors of the preimage.  And the forms you want to preserve are also bilinear! So if we look at $B_{1}(u,v)$, with $u$ and $v$ in $C$, we can write $u = a_{1}X_{1} + \ldots + a_{n}X_{n}$ and $v = b_{1}X_{1} + \ldots + b_{n}X_{n}$.  Then
$$B_{2}(F(u),F(v)) = \sum_{i}\sum_{j} a_{i}b_{j}B_{2}(F(X_{i}),F(X_{j})) = \sum_{i}\sum_{j} a_{i}b_{j}B_{1}(X_{i},X_{j}) = B_{1}(u,v).$$
(Note that we have two different bilinear forms, $B_{1}$ on $C$ and $B_{2}$ on $D$.)
