I'm studying inner product spaces, and I had this question.
Is my generalization of this private cases of added elements into a sum correct, and if so, is my explanation of the transition from the private case to the sum "enough", formally, or do I need to prove it inductively?
I don't think that I do, but I'm not entirely sure.
Here's the question:
$ \text{Let} \> V \> \text{be a vector space} \text{over field} \> F. \\ \text{Let} \> v_1,\dots,v_n \in V. \\ \text{Let} \> \xi_1,\dots,\xi_i,\eta_1,\dots,\eta_j \in F. \\ \text{Prove:} \> \left\langle \sum_{i=1}^{n}\xi_{i}u_{i},\sum_{j=1}^{m}\eta_{j}v_{j}\right\rangle =\sum_{i=1}^{n}\sum_{j=1}^{m}\xi_{i}\overline{\eta_{j}}\left\langle u_{i},v_{j}\right\rangle. $
This is what I did:
$ \left\langle \xi_{1}u_{1}+\xi_{2}u_{2},\eta_{1}v_{1}+\eta_{2}v_{2}\right\rangle =\left\langle \xi_{1}u_{1},\eta_{1}v_{1}+\eta_{2}v_{2}\right\rangle +\left\langle \xi_{2}u_{2},\eta_{1}v_{1}+\eta_{2}v_{2}\right\rangle \because\text{linearity} =\overline{\left\langle \eta_{1}v_{1}+\eta_{2}v_{2},\xi_{1}u_{1}\right\rangle }+\overline{\left\langle \eta_{1}v_{1}+\eta_{2}v_{2},\xi_{2}u_{2}\right\rangle }\because\text{hermiticity} \\ =\overline{\left\langle \eta_{1}v_{1},\xi_{1}u_{1}\right\rangle }+\overline{\left\langle \eta_{2}v_{2},\xi_{1}u_{1}\right\rangle }+\overline{\left\langle \eta_{1}v_{1},\xi_{2}u_{2}\right\rangle }+\overline{\left\langle \eta_{2}v_{2},\xi_{2}u_{2}\right\rangle }\because\text{linearity}\\ =\overline{\eta_{1}\left\langle v_{1},\xi_{1}u_{1}\right\rangle }+\overline{\eta_{2}\left\langle v_{2},\xi_{1}u_{1}\right\rangle }+\overline{\eta_{1}\left\langle v_{1},\xi_{2}u_{2}\right\rangle }+\overline{\eta_{2}\left\langle v_{2},\xi_{2}u_{2}\right\rangle }\because\text{homogeneity}\\ =\overline{\eta_{1}}\overline{\left\langle v_{1},\xi_{1}u_{1}\right\rangle }+\overline{\eta_{2}}\overline{\left\langle v_{2},\xi_{1}u_{1}\right\rangle }+\overline{\eta_{1}}\overline{\left\langle v_{1},\xi_{2}u_{2}\right\rangle }+\overline{\eta_{2}}\overline{\left\langle v_{2},\xi_{2}u_{2}\right\rangle }\because\text{commutativity of}\ \cdot_{\mathbb{C}} =\overline{\eta_{1}}\overline{\overline{\left\langle \xi_{1}u_{1},v_{1}\right\rangle }}+\overline{\eta_{2}}\overline{\overline{\left\langle \xi_{1}u_{1},v_{2}\right\rangle }}+\overline{\eta_{1}}\overline{\overline{\left\langle \xi_{2}u_{2},v_{1}\right\rangle }}+\overline{\eta_{2}}\overline{\overline{\left\langle \xi_{2}u_{2},v_{2}\right\rangle }}\because\text{hermiticity}\\=\overline{\eta_{1}}\left\langle \xi_{1}u_{1},v_{1}\right\rangle +\overline{\eta_{2}}\left\langle \xi_{1}u_{1},v_{2}\right\rangle +\overline{\eta_{1}}\left\langle \xi_{2}u_{2},v_{1}\right\rangle +\overline{\eta_{2}}\left\langle \xi_{2}u_{2}v_{2}\right\rangle \because\forall z\in\mathbb{C}:\overline{\overline{z}}=z =\overline{\eta_{1}}\xi_{1}\left\langle u_{1},v_{1}\right\rangle +\overline{\eta_{2}}\xi_{1}\left\langle u_{1},v_{2}\right\rangle +\overline{\eta_{1}}\xi_{2}\left\langle u_{2},v_{1}\right\rangle +\overline{\eta_{2}}\xi_{2}\left\langle u_{2},v_{2}\right\rangle \because\text{linearity}\\ =\xi_{1}\overline{\eta_{1}}\left\langle u_{1},v_{1}\right\rangle +\xi_{1}\overline{\eta_{2}}\left\langle u_{1},v_{2}\right\rangle +\xi_{2}\overline{\eta_{1}}\left\langle u_{2},v_{1}\right\rangle +\xi_{2}\overline{\eta_{2}}\left\langle u_{2},v_{2}\right\rangle \because\text{commutativity of}\ \mathbb{\cdot_{\mathbb{C}}} \Rightarrow\left\langle \xi_{1}u_{1}+\xi_{2}u_{2},\eta_{1}v_{1}+\eta_{2}v_{2}\right\rangle =\sum_{i=1}^{n}\sum_{j=1}^{m}\xi_{i}\overline{\eta_{j}}\left\langle v_{i}v_{j}\right\rangle $
Note: $m,n$ are not given. I took a private case of $m=2,n=2$ and tried generalizing it into a sum.
I'd appreciate remarks a lot
Your equations might be right, they are to big and not readable. You seem to have the right idea, however, look at it yourself: Would you appreciate such a proof? Try to get rid of the long equation, don't be afraid to add some text, you don't have to squeeze in the explanations next to the equations.
The last row is of course wrong, you don't have equality between the left side with $m=n=2$ and the general case on the right side (except, of course, for $m=n=2$...^^). So here a proof by induction would come in handy.