Why is linearity not satisfied for this inner product?

Question

The inner product given is $\langle u,v \rangle = u_{1}^{2}v_{1}^{2} + u_{2}^{2}v_{2}^{2}$. If I am thinking of linearity correctly, then all that means is the coefficients in front of each like term would be the same, so there wouldn't be something like $5u_{1}^{2}$ and $3u_{1}^{2}$. So why doesn't this inner product work?

Edit: In my notes, linearity is written as $\langle u,c_{1}v_{1} + c_{2}v_{2}\rangle$ = $c_{1}\langle u, v_{1} \rangle + c_{2}\langle u, v_{2} \rangle$. I still don't get it. Does linearity mean you can't have terms that are to the third power and above? If so, why?

Answer 1

To make things simple, try an example with $c_1=c_2=1$.

$$\begin{eqnarray*}\langle u,v + w\rangle & = & u_{1}^{\,\,2}(v+w)_{1}^{\,\,2} + u_{2}^{\,\,2}(v+w)_{2}^{\,\,2}\\ & = & u_{1}^{\,\,2}(v_1+w_1)^{2} + u_{2}^{\,\,2}(v_2+w_2)^{2}\\\langle u, v \rangle + \langle u, w \rangle & = & u_{1}^{\,\,2}v_{1}^{\,\,2} + u_{2}^{\,,2}v_{2}^{\,\,2} + u_{1}^{\,\,2}w_{1}^{\,\,2} + u_{2}^{\,\,2}w_{2}^{\,\,2}\\& = & u_{1}^{\,\,2}\left(v_{1}^{\,\,2} + w_{1}^{\,\,2}\right)+ u_{2}^{\,\,2}\left(v_{2}^{\,\,2} + w_{2}^{\,\,2}\right)\end{eqnarray*}$$These aren't the same thing.

Answer 2

What you quote as written in your notes is the correct definition. It does indeed follow from that definition that

you can't have terms that are to the third power and above.

You ask "why?". The answer is that with powers greater than $1$ the identity $$ \langle u,c_{1}v_{1} + c_{2}v_{2}\rangle = c_{1}\langle u, v_{1} \rangle + c_{2}\langle u, v_{2} \rangle $$ will fail. Try it with the example in your question.