Unclear why solving part of the vector projection gives the change of base vector In the change of basis section of the linear algebra track on coursera the professor solves
$$\frac{r_e\cdot b_1}{\left|b_1\right|^2}$$
with $r$ being a vector and $b_1$ and $b_2$ being the new basis. He later solves the full vector projection equation
$$\frac{r_e\cdot b_1}{\left|b_1\right|^2}*b_1$$
However, he demonstrates solving this part of the vector projection equation $\frac{r_e\cdot b_1}{\left|b_1\right|^2}$ gives one of the $2$ components of the new vector in the context of the changed bases, once you do the same for the projection on the other basis you get the new vector in the context of the changed basis. What is the intuition behind this?
 A: We are projecting the vector $r_e$ onto the new basis vector $b_1$ and $b_2$ and in the new basis the coordinate for the vector $r_e$ are
$$r'_e=\left(\frac{r_{e} . b_{1}}{\left | b_{1} \right |},\frac{r_{e} . b_{2}}{\left | b_{2} \right |}\right)$$
and
$$r'_e=\frac{r_{e} . b_{1}}{\left | b_{1} \right |} \frac{b_1}{\left | b_{1} \right |}+\frac{r_{e} . b_{2}}{\left | b_{2} \right |}\frac{b_2}{\left | b_{2} \right |}=\frac{r_{e} . b_{1}}{\left | b_{1} \right |^{2}}b_1+\frac{r_{e} . b_{2}}{\left | b_{2} \right |^{2}}b_2$$
More in general if $b_i$ are  th enew vector basis we have that


*

*$r_{e} \cdot\frac{ b_{i}}{\left | b_{i} \right |}$ is the scalar projection of $r_e$ onto $b_i$ (it is a number)

*$\left(r_{e} \cdot \frac{ b_{i}}{\left | b_{i} \right |}\right)\frac{ b_{i}}{\left | b_{i} \right |}$ is the vector projection of $r_e$ onto $b_i$ (it is a vector)
Note that in the new basis the vector $r'_e$ is indicated by the new components onto $b_i$ that is
$$r_e'=\left(\ldots,r_{e} \cdot\frac{ b_{i}}{\left | b_{i} \right |},\ldots\right)$$
