"All scalars are invariant": meaning in the context of changing basis I'm interested in understanding a comment by Chapter 0 of McCullaugh (1987). To keep the question self-contained, I'll provide some context below. Thank you.

Suppose I have vector space $V$ with $\{e_1,\ldots,e_n\}$. Let $v\in V$ be represented by $\sum_{i=1}^nv_ie_i$ where $v_i$'s are scalars. Let $K$ be some invertible $n\times n$ matrix and define new basis vectors
$$
\bar{e}_i=\sum_{j=1}^nK_{ij} e_j\iff (e_1\;\cdots\;e_n)K'=(\bar{e}_1\;\cdots\;\bar{e}_n).
$$
Then $v$ can also be written as $\sum_{i=1}^n\bar{v}_i\bar{e}_i$ where
$$
\begin{pmatrix}
\bar{v}_1 \\ \vdots \\ \bar{v}_n
\end{pmatrix}=(K')^{-1}\begin{pmatrix}
v_1 \\ \vdots \\ v_n
\end{pmatrix}.
$$
Comment of interest (first paragraph, page 14): 

...(that) basis vectors are arbitrary up to linear transformation effectively means that all calculations are independent of the choice of the basis. In particular, all scalars are invariant i.e. independent of the choice of basis.

Question: what does invariant mean here? The new coordinates $\bar{v}_1,\ldots,\bar{v}_n$ above are not the same as the original coordinates $v_1,\ldots,v_n$ which are scalars.
 A: The comment as it is in your post is taken out of context: in your post, it seems to be a part of some considerations about vectors - in which case one naturally asks how could scalars be particular cases of vectors (exactly your question). In reality, this comment in its original context is a part of some considerations about tensors, and in this case things are clear, because a scalar means a tensor of type $(0,0)$ or, if you prefer, an element of $V^0 \otimes (V^*)^0$, which is after all isomorphic to the base field.
To clarify this with an example, consider $\Bbb R^2$ with the basis $\{(1,0), (0,1)\}$ and the scalar $3 \in \Bbb R$. Now perform some arbitrary change of basis in $\Bbb R^2$ - it doesn't matter which one. Does $3$ change in any way? No, it doesn't even feel this change of basis - and this is the simplest type of invariance under a change of coordinates: being constant. (The reason why scalars -i.e. numbers- are invariant under such changes of basis being that they do not, in fact, have coordinates. Think for yourself: what are the $(x,y)$ coordinates of the number $3$?)
A: Of course, writing "all calculations" is misleading. What one can say is the following: Given two vectors $x$, $y\in V$ the sum $z:=x+y$ is a well defined element of $V$ which does not depend on the chosen basis. In the same vein: You obtain the the coordinates of $z$ by adding the coordinates of $x$ and of $y$, whatever the chosen basis is. Similarly, the vector $w:=\alpha x$ is a well defined element of $V$ which does not depend on the chosen basis. Furthermore you obtain the the coordinates of $w$ by multiplying the coordinates of $x$ with this very $\alpha$, whatever the chosen basis is.
