TITLE: What is a part of a vector called?
QUESTION [0 upvotes]: Assume we have $x \in R^3$, which means we have $x = [x_1, x_2, x_3]^T$.
Then we have another $\bar{x}=[x_1,x_3]$. It is just the $x$ without $x_2$, and $x_1, x_3$ have the same meaning in both $\bar{x}$ and $x$.
My question is what is the relationship between $\bar{x}$ and $x$?
Aka. We can say that $\bar{x}$ is the ? of $x$.

REPLY [1 votes]: Arguably, there is no meaningful relation. To counter the claim that such is a projection, note that you could find such a $\pi$ almost regardless of your original vectors.
Let me rephrase/generalize your idea in terms of standard linear algebra terminology. We start with two vector spaces $\mathbb{R}^{n}$ and $\mathbb{R}^{m}$ with $n > m$ (in your case, $n = 3, m = 2$). Take some bases $\{ e_{i} : 1 \leq i \leq n \}$ and $\{ f_{i} : 1 \leq i \leq m \}$; in your case you take the standard bases.
Then you have two vectors
  $$ x = \sum_{i = 1}^{n} x_{i} e_{i} $$
and
  $$ \bar{x} = \sum_{i = 1}^{m} x_{i} f_{i}. $$
I admit technically you have $\bar{x}$ with coordinates $(x_{1}, x_{3})$ rather than $(x_{1}, x_{2})$ but hopefully it is clear that this is essentially the same.
Then the claim is that you have a 'natural' linear map $\pi: \mathbb{R}^{n} \to \mathbb{R}^{m}$ with $\pi(x) = \bar{x}$. Certainly, you could take
  $$ \pi: e_{i} \mapsto
       \begin{cases}
         f_{i} \ & \mathrm{if} \ i \leq m \\
         0 \ & \mathrm{otherwise}
       \end{cases}.
  $$
However, note that this is dependent on the bases you chose in the first place. Given any two nonzero vectors $u \in \mathbb{R}^{n}$ and $v \in \mathbb{R}^{m}$ you can find bases for which $\pi(u) = v$ when $\pi$ is defined this way. Basically, this means that if $\bar{x}$ is a 'projection' of $x$, then any nonzero vector in $\mathbb{R}^{m}$ is a 'projection' of any nonzero vector in $\mathbb{R}^{n}$.
To see this, note that we can construct two bases with $u = e_{1}$ and $v = f_{1}$. I'm not sure if there is a simple way to show this, but it is a standard result that any linearly independent set of vectors can be extended to a basis ($\{u\}$ is linearly independent in $\mathbb{R}^{n}$, and $\{v\}$ is a linearly independent set in $\mathbb{R}^{m}$).
Furthermore, to make the situation the same, we let $x_{1} = 1$ and $x_{i} = 0$ for all $i > 1$. This gives $u = \sum x_{i} e_{i}$ and $v = \sum x_{i} f_{i}$, and $\pi(u) = v$.
The central point in all of this is that no basis is special. It may feel like the standard basis of $\mathbb{R}^{n}$ is special, but vector spaces are defined in such a way to make this not true. That is, vector spaces are defined in a way that emphasizes the fact that so much in linear algebra is completely unaffected by our choice of basis (which is very convenient for most applications).
One caveat is that this is not quite true if you are talking about inner product spaces. Specifying an inner product makes some bases special (namely the orthonormal ones).
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