Understanding Vector Spaces well I really want to get a strong grasp of abstract linear algebra for 2 reasons, I want to go deeper in pure mathematics for joy (abstract algebra, number theory etc) and to have a deeper understanding of how machine learning algorithms work under the hood.
I have been struggling with understanding vector spaces well. I have read through the axioms, their proofs and other examples/sample exercises on them in both textbooks as well as here on math stack exchange. When I look at the solutions both here and on some math stack exchange on why a certain object is a vector space or not, I don't follow the reasoning. Here is one such old question from here (this is just 1 example, there are many others I don't follow)
Why does vector sum $(x_1,x_2)+'(y_1,y_2)=(x_1+2y_1, 3x_2-y_2)$ and $(cx_1,cx_2)$ fail to hold the axiom of vector space?
When I looked at the very brief accepted answer that just states:
Let =(1,0) and =(0,1). Then +′=(1,−1) and +′=(2,3).
I don't understand how did they come up with (2,3)?
I looked at the 2nd solution which has a lot more detail and am quiet confused. It seems like the algebraic manipulations are done in a different way from what I would expect knowing high school algebra?
For example here (sorry still need to learn latex):
(+)+=(1+21,32−2)+(1,2)=(1+21+21,3(32−2)−2)=(1+21+21,92−32−2)
could someone explain to me how each of these equations are arrived at it detail? I don't understand why you can add a "+2z1" when it was just "z1" by itself? Also, in the following equation why can you do "3(32−2)" it seems like you are applying the 3 multiplier 2 times over? And the same reasoning I am missing from +(+) equation but I guess it would be similar.
Thank you!
 A: Normal vector addition is defined so that, given two (two-dimensional) vectors $\vec x = (x_1, x_2)$ and $\vec y = (y_1, y_2)$, their sum $\vec x + \vec y = (x_1, x_2) + (y_1, y_2)$ is defined to be equal to $(x_1 + y_1,\ x_2 + y_2)$. In other words, $$(x_1, x_2) + (y_1, y_2) \overset{\rm def}{=} (x_1 + y_1,\ x_2 + y_2).$$
(Note that the $+$ sign on the left side of the $\overset{\rm def}{=}$ symbol is really a new operation that we're defining, since we haven't previously defined what it means to add two vectors together.  The two $+$ signs on the right, however, simply denote the ordinary addition of two numbers.)
Now, the question asks what would happen if we were to define this new vector addition operator differently.  To avoid confusing this alternative definition with the normal one given above, let's denote the alternative addition operator by a different symbol.  The question uses $+'$ for this new operator, but I don't really like that choice for typographical reasons (the apostrophe looks too disconnected from the plus sign), so let me call it $\oplus$ instead.  Thus, our alternative vector addition rule now looks like this: $$(x_1, x_2) \oplus (y_1, y_2) \overset{\rm def}{=} (x_1 + 2y_1,\ 3x_2 - y_2).$$
Now, the question is whether the set $\mathbb R^2$, equipped with the usual vector multiplication operator $\cdot$ and this new alternative addition operator $\oplus$, could also satisfy the definition of a vector space over $\mathbb R$.
As it turns out, the answer is "no."  And to prove that, all we need to do is give at least one example of vectors for which at least one of the axioms in the definition of a vector space fails to hold.
Now, one of these axioms is that vector addition needs to be commutative: if $+$ denotes the addition operator in a vector space, then $\vec x + \vec y$ needs to always be equal to $\vec y + \vec x$.  Clearly this axiom does hold for the usual addition operator, as defined at the top of this answer.  Does it also hold for the alternative operator $\oplus$?
A moment's thought should reveal that it does not, since $x_1 + 2y_1$ is generally not equal to $y_1 + 2x_1$.  (In fact, the two expressions are only equal if $x_1 = y_1$.)  And, for that matter, $3x_2 - y_2$ is also not equal to $3y_2 - x_2$ unless $x_2 = y_2$.
So this alternative addition operator $\oplus$ in fact fails the commutativity requirement just about as badly as it's possible to fail it: $\vec x \oplus \vec y \ne \vec y \oplus \vec x$ whenever $\vec x \ne \vec y$.
But, as I noted earlier, we don't actually need to prove that this commutativity failure occurs for all pairs of non-equal vectors $\vec x$ and $\vec y$.  All we need to do is show one pair of vectors for which it fails to hold.  So let's just pick any two distinct vectors — say, $\vec x = (0,1)$ and $\vec y = (1,0)$ — and do the arithmetic:
$$\begin{alignat}{3}
\vec x \oplus \vec y &=\;& (0,1) \oplus (1,0) &=\;& (0 + 2 \cdot 1,\ 3 \cdot 1 - 0) &= (2,3) \\
\vec y \oplus \vec x &=\;& (1,0) \oplus (0,1) &=\;& (1 + 2 \cdot 0,\ 3 \cdot 0 - 1) &= (1,-1).
\end{alignat}$$
Of course, you could just as well pick any other pair of distinct vectors.  And of course it doesn't matter whether you call the example vectors $\vec x$ and $\vec y$ or $\vec u$ and $\vec v$ or whatever.  And it also doesn't really matter whether you write your vector symbols as $\vec x$ or $\bar x$ or $\mathbf x$ or just $x$, although you should generally try to follow whichever convention your textbook uses.
