Is the Gram-Schmidt procedure a bijection?

Question

Since the Gram-Schmidt procedure yields a single orthonormal output for a given linearly independent input, we can view the procedure as a function which maps an original linearly independent list of vectors in finite dimensional vector space $V$ to an orthonormal list of vectors in $V$ that span the same vector space as the span of the original list.

In other words, define

$$G:\{\text{all linearly independent lists of vectors in } V\} \xrightarrow{} \{\text{all orthonormal lists of vectors in } V\}$$

as

$$(v_1,...,v_k) \mapsto (e_1,...,e_k)$$

such that $(e_1,...,e_k)$ is the orthonormal output of the Gram-Schmidt procedure, so span$(v_1,...,v_k)$ = span$(e_1,...,e_k)$.

My first question: is this function injective? I.e. can two distinct linearly independent lists (necessarily with the same span) yield the same orthonormal basis after applying Gram-Schmidt?

My second question: is $G$ surjective? I.e. given an orthonormal list of vectors in $V$, can I construct a linearly independent list of vectors in $V$ that will yield my orthonormal list after applying Gram-Schmidt?

My intuition is that $G$ is not injective because the set of all orthonormal lists is a proper subset of the set of all linearly independent lists, so the codomain is a proper subset of the domain. I feel like $G$ should be surjective but do not know how to prove this.

Answer 1

Thanks to Michael Burr's comment, I see that the function is surjective, because for any orthonormal list $(e_1,...,e_k)$ of vectors in $V$, $(e_1,...,e_k)$ is linearly independent and $G(e_1,...,e_k)=(e_1,...,e_k)$.

And the function is not injective. For example, $G(e_1) = e_1$ and $G(2e_1)=e_1$.

Answer 2

Your intuition about the first question is right.

I'll show you example in 3D-space. Most important aspect of Gram-Schmidt procedure is that the for the new basis produced the span of the first $k$ vectors is the same as the span of the first $k$ vectors of the original basis, for all $k$.

This shows that the last vector in the new basis is in the 1-dimensional orthogonal complement of the $n-1$ vectors constructed, and so will be independent of the last basis vector of the original basis.

To illustrate with a 3-dimensional example let us take a vector $v_2, v_3 v\in \mathbf{R}^3$ with $v_2$ in the span of $e_1,e_2,$ while $v_3$ not in that span. More precisely $v_2= (x,y,0), y\ne0, v_3= (a,b,c)$ with arbitrary $a,b,c, c\ne 0$. For all the choices of $v_2,v_3$ satisfying the above conditions with $v_1=e_1, v_2, v_3$ will be a basis and Gram-Schmidt will yield the standard basis $e_1,e_2,e_3$ as the output.

Use this trick to show surjectivity to construct a basis yield a desirable onb as output.