Change of basis vector I don't have a problem, I just need help with the theory. I feel like I keep misunderstanding things when it comes to solving problems that involves changing from a basis to another.
Lets say we are given 3 linearly independent vectors $v_1, v_2, v_3$. If we are to build a basis of these vectors, we would put them in a bracket right, but they are said to be a new basis that has it's coordinates with respect to the standard basis?
But what if we wanted to write these new basis with its own coordinates? Also how can we write the standard basis with respect to this new basis?
I'd really appreciate it if someone can help me out, or give me a YT video recommendation or another website. I would also really appreciate it if someone could give me questions to work with to understand it better. I tried googling myself but I only found what I already know, which is how to change from one base to another etc.
 A: Let $V$ be a vector space with a basis $\mathcal{B}_s=\{e_1,e_2,...,e_n\}$ called the standard basis. Then any vector $P\in V$ of this vector space is in the form $$P_s=(x_1,x_2,...,x_n)=x_1e_1+x_2e_2+...+x_ne_n$$ and this form is called standard coordinates of $P$. I denoted by a subscript $s$, but we may omit it when it is clear that it is standard coordinates. Then it is easy to see that $(e_1)_s=(1,0,...,0)$, $(e_2)_s=(0,1,0,...,0)$,... etc.
Let us now be given another basis $\mathcal{B}_y=\{v_1,v_2,...,v_n\}$ where $y$ is for "yeni" meaning "new" in Turkish. I had to do this. I used $n$. Then any vector $P\in V$ of this vector space is in the form $$P_y=(x_1',x_2',...,x_n')=x_1'v_1+x_2'v_2+...+x_n'v_n$$ and this form is called new  cordinates of $P$. Then it is easy to see that $(v_1)_y=(1,0,...,0)$, $(v_2)_y=(0,1,0,...,0)$,... etc. Notice that, this is independent of the old (standard) basis.
The elements of the new basis can be given in terms of the standard basis: Omitting the subscript $s$, let $v_1=(v_{11},v_{12},...,v_{1n})$, $v_2=(v_{21},v_{22},...,v_{2n})$,...etc. in standard coordinates.
Main question: Given $P_s=(x_1,x_2,...,x_n)$, how can we find $P_y=(x_1',x_2',...,x_n')$?
Answer: We treat the vectors as column matrices! And we form the key matrix below
$$A=\left[\begin{matrix}v_1|v_2|...|v_n\end{matrix}\right]
=\left[\begin{matrix}v_{11}&v_{21}&...&v_{n1}\\v_{12}&v_{22}&...&v_{n2}\\..&..&&..\\v_{1n}&v_{2n}&...&v_{nn}\\\end{matrix}\right].$$
Then, $P_y=A^{-1}P_s$. Proof is left to the reader.
Standard basis vectors now can be found in terms of new basis, that is, with new coordinates like any vector can be: $(e_1)_y=A^{-1}(e_1)_s$, $(e_2)_y=A^{-1}(e_2)_s$,...etc. It turns out that these are the columns of $A^{-1}$.
Question: Let $\mathcal{B}_y=\{(1,1),(1,2)\}$. If $P=(3,2)$, find $P_y$. Also, find the standard basis vectors with new coordinates.
Answer: Form $A=\left[\begin{matrix}v_1|v_2\end{matrix}\right]=\left[\begin{matrix}1&1\\1&2\end{matrix}\right]$. Then, $A^{-1}=\left[\begin{matrix}2&-1\\-1&1\end{matrix}\right]$ and $P_y=A^{-1}P_s=\left[\begin{matrix}2&-1\\-1&1\end{matrix}\right]\left[\begin{matrix}3\\2\end{matrix}\right]=\left[\begin{matrix}4\\-1\end{matrix}\right]$. Also, $(e_1)_y=\left[\begin{matrix}2&-1\\-1&1\end{matrix}\right]\left[\begin{matrix}1\\0\end{matrix}\right]=\left[\begin{matrix}2\\-1\end{matrix}\right]$ and $(e_2)_y=\left[\begin{matrix}2&-1\\-1&1\end{matrix}\right]\left[\begin{matrix}0\\1\end{matrix}\right]=\left[\begin{matrix}-1\\1\end{matrix}\right]$.
A: If $v_1,v_2,v_3$ are going to be a new basis it is because they must be  linearly independent.
If they already are linearly independent then they are used to write any other vector, say $w$ as
$$w=c^1v_1+c^2v_2+c^3v_3,$$
for some scalar $c^i$. One says that the $c^i$ are the $w$ components in that basis.
So for them
$$v_1=1v_1+0v_2+0v_3,$$
$$v_2=0v_1+1v_2+0v_3,$$
$$v_3=0v_1+0v_2+1v_3,$$
where you can see their components with respect themselves.
Now if
$$v_1=a_{11}e_1+a_{21}e_2+a_{31}e_3,$$
$$v_2=a_{12}e_1+a_{22}e_2+a_{32}e_3,$$
$$v_3=a_{13}e_1+a_{23}e_2+a_{33}e_3,$$
then one associates the matrix
$$A=
\left(
\begin{array}{ccc}
a_{11}&a_{12}&a_{13}\\
a_{21}&a_{22}&a_{23}\\
a_{31}&a_{32}&a_{33}.
\end{array}
\right)$$
Since the vectors $v_i$ are linearly independent this matrix is nonsingular and the has inverse. It's  going to happen that
$$e_1=b_{11}v_1+b_{21}v_2+b_{31}v_3,$$
$$e_2=b_{12}v_1+b_{22}v_2+b_{32}v_3,$$
$$e_3=b_{13}v_1+b_{23}v_2+b_{33}v_3,$$
where the corresponding associated matrix $B$ is $A^{-1}$.
When studying the subject it is worthwhile to have access to several textbooks to achieve a contrasting view, which is better than the boring YT explains.
