For example, when finding change of bases $P_{a \leftarrow b}$ such that it goes from polynomial of degree one to polynomial of degree one, $P_1 \rightarrow P_1$, then I put the polynomial in column vector form for the augmented matrix instead of row vectors. For $a=[2+x,3+4x]$ & $b=[9+6x, 8+7x]$, why is this augmented matrix correct:
$$
\begin{bmatrix}
2 & 3 &|& 9 & 8 \\
1 & 4 &|& 6 & 7
\end{bmatrix}
\quad
$$
while this, my original method, is incorrect:
$$
\begin{bmatrix}
2 & 1 &|& 9 & 6 \\
3 & 4 &|& 8 & 7
\end{bmatrix}
\quad
$$
I know I reduce the left side to the identity and the right side then becomes my transition matrix.
Best Answer
Let $V$ be a finite dimensional vector space and $e_i$ be the elements of one basis of $V$ and let $f_i$ be the elements of another basis. The change of basis matrix from $e_i$ to $f_i$ has as its entries the scalars $a_{ij}$ such that $f_i = \sum_j a_{ij}e_j$. If you associate $e_i$ with the standard basis on $\mathbb{R}^n$, then the columns of the matrix will be the coordinate representation of the new basis in $\mathbb{R}^n$.
For example, with the polynomials you mention we have \begin{aligned} e_1 &= 2+x\\ e_2 &= 3+4x\\ \end{aligned} and \begin{aligned} f_1 &= 9+6x\\ f_2 &= 8+7x.\\ \end{aligned} since \begin{aligned} f_1 &= 3.6 e_1 + 0.6 e_2\\ f_2 &= 2.2 e_1 + 1.2 e_2, \end{aligned} (please check my work to make sure there are no fat-finger-induced Matlab typing errors...) it follows that the change of basis matrix is \begin{aligned} A = \begin{bmatrix} 3.6 & 0.6\\ 2.2 & 1.2\\ \end{bmatrix}. \end{aligned}
Now: why the first augmented matrix and not the second?
Note that to find the matrix above you need to solve the system \begin{aligned} a_{11}(2+x) + a_{12}(3+4x) &= 9 + 6x\\ a_{21}(2+x) + a_{22}(3+4x) &= 8 + 7x.\\ \end{aligned} Since $1$ and $x$ are a basis of $P_1(\mathbb{R})$, they are linearly independent, so we can rewrite this as \begin{aligned} 2a_{11} + 3a_{12} &= 9\\ a_{11} + 4a_{12} &= 6\\ 2a_{21} + 3a_{22} &= 8\\ a_{21} + 4a_{22} &= 7,\\ \end{aligned} which is equivalent to the simultaneous equations \begin{aligned} \begin{bmatrix} 2 & 3\\ 1 & 4\\ \end{bmatrix} \begin{bmatrix} a_{11}\\ a_{12}\\ \end{bmatrix} = \begin{bmatrix} 9\\ 6\\ \end{bmatrix} \end{aligned} and \begin{aligned} \begin{bmatrix} 2 & 3\\ 1 & 4\\ \end{bmatrix} \begin{bmatrix} a_{21}\\ a_{22}\\ \end{bmatrix} = \begin{bmatrix} 8\\ 7\\ \end{bmatrix}. \end{aligned} These two lead to the first augmented matrix rather than the second.