Answer
see the counterexample below.
Work Step by Step
Since $\langle (u_1,u_2), (v_1, v_2 )\rangle= u_1v_1 $, let $u=(0,1)$, then $\langle u.u\rangle=\langle (0,1).(0,1)\rangle=0$. But $u\neq (0,0)$, so the function does not define an inner product.