Finding a change of basis matrix when the basis vectors of one basis are in terms of the other I'm trying to solve this problem. 
V is a vector space with bases:
$$B = \lbrace v_1, v_2, v_3\rbrace\quad \text{and}\quad B' = \lbrace w_1, w_2, w_3\rbrace.$$
Assume: 
\begin{align*}
v_1 &= w_1 + w_2 + 3w_3 \\
v_2 &= 2w_1 + 3w_2 + 2w_3\\
v_3 &= 3w_1 + 4w_2 + 4w_3
\end{align*}
Intuitively, I think that this represents the matrix equation:
$$
\begin{bmatrix} 1&1&3\\2&3&2 \\ 3&4&4\\ \end{bmatrix}
\begin{bmatrix} w_1\\w_2\\w_3\\ \end{bmatrix} = 
\begin{bmatrix} v_1\\v_2\\v_3\\ \end{bmatrix}
$$
And therefore the matrix $$
\begin{bmatrix} 1&1&3\\2&3&2 \\ 3&4&4\\ \end{bmatrix}$$ is the transformation matrix from $B' \to B$.
However by the definition of a transformation matrix (columns being the basis vectors), the matrix should be the transpose $$
\begin{bmatrix} 1&2&3\\1&3&4 \\ 3&2&4\\ \end{bmatrix}.$$
Which of these is right? And why?
Thank you!
 A: Let's revisit the theorem relating to change of basis once again.We will assume the field to be $\mathbb{R}$

Suppose $P$ is an $n\times n$ invertible matrix over $\mathbb{R}.$ Let $V$ be an $n$-dimensional vector space over $\mathbb{R},$ and let $\scr C$ be an ordered basis of $V.$ Then there is a unique ordered basis $\scr D$ of $V$ such that 
\begin{align}
\tag{1}[\alpha]_{\scr C} = P[\alpha]_{\scr D} 
\end{align}
  for every vector $\alpha \in V.$ 
$\Big($ $[\alpha]_{\scr B}$ is the $\color{red}{\text{coordinate matrix}}$ of $\alpha$ relative to the ordered basis $\scr B.$ $\Big)$

Here the columns of $P$ are the $\color{red}{\text{coordinates}}$ of the basis vectors of $\scr D$ with respect to the ordered basis $\scr C.$ 
So note that $\color{red}{\text{coordinate matrices}}$ in your case will always be $3\times 1$ matrices with entries from $\mathbb{R}.$
Suppose in your case I consider the ordered basis $B.$ A vector $\alpha \in V$ with coordinate matrix $\begin{bmatrix} 2\\5\\3\\ \end{bmatrix}$ means $\alpha =2v_1+5v_2+3v_3. $
The matrix equation you wrote has NO coordinate matrices.
