Is the sum over a dot product, the same as a sum of a vector dotted with the second vector

Question

Is the following statement true?

$\sum_i\left(\vec{a}_i \cdot \vec{b}\right)=\left(\sum_i \vec{a}_i\right) \cdot \vec{b}$

I believe the answer is yes. More importantly, how can I go about proving it. Maybe there's some well known answer out there to link to, but I would really be curious if someone could simply point me in the right direction of thought to come up with a proof on my own. One can come up with similar questions about cross products as well.

Answer 1

After only a small amount more thought, and while considering how we prove something like $\sum_i(c_i\cdot d)=\sum_i(c_i)*d$ and the answer actually seems pretty trivial. At least for this case, you can just imagine writing out $c_1\cdot d+c_2\cdot d+c_3\cdot d+...=(c_1+c_2+c_3...) \cdot d$. You can similarly do the same thing with dot products (or cross products) and just factor out the constant (vector).

Answer 2

Finite sums you can interchange without much thought.

$\sum_i\sum_ja_{ij}b_j=\sum_jb_j\left(\sum_ia_{ij}\right)$