Imagine I have an arbitrary vector $\mathbf{v}\in\mathbb{R}^n$ and say it can be represented as
$$
\mathbf{v}=(v_1,…,v_n)^T
$$
where it's understood that the indexes are in increasing order. Imagine now the vector $\mathbf{u}\in\mathbb{R}^{n-1}$ obtained by simply removing entry with index $i$ of $\mathbf{v}$. What's the best (and most "economic") way to represent such vector? My idea was to simply interpret the vector as a sequence and write
$$
\mathbf{u}=\{v_j\}_{j\neq i}^T
$$
but I'm not sure whether this is correct and/or there is a better way of writing it. Any ideas?
Edit: I'm aware that
$$
\mathbf{u}=(v_1,…,v_{i-1},v_{i+1},…,v_n)^T
$$
can be used, but I was wondering whether a more "elegant" representation could be possible. If $i=1$ or $i=n$, in this notation, it is just strange that we included $v_1$ or $v_n$ in $\mathbf{u}$ in the first place. Purely an aesthetic preference, I know, but still.
Best Answer
If without writing any components perhaps for $\mathbf v \in \mathbb R^n$ you could write $\mathbf u = \mathbf v/e_i \in \mathbb R^{n-1}$, as a ''qoutient'', where $e_i$ is the $i$-th standard basis vector in $\mathbb R^n$. I hope you know quotient spaces.
In some textbooks hats indicate the omission of elements $$ \mathbb R^{n}/\text{span}\{e_i\} \cong \mathbb R^{n-1} \ni \mathbf v/e_i = (v_1,\dots,\widehat{v_i},\dots,v_n)^T = (v_1,\dots,v_{i-1},v_{i+1},\dots,v_n)^T $$