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!
Let's revisit the theorem relating to change of basis once again.We will assume the field to be $\mathbb{R}$
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.