1. **Problem:** Determine if the function $\langle u, v \rangle = 3u_1v_1 + 5u_2v_2$ defines a valid inner product on $\mathbb{R}^2$. 2. **Recall the axioms of an inner product:** - Linearity in the first argument: $\langle au + bw, v \rangle = a\langle u, v \rangle + b\langle w, v \rangle$ - Symmetry: $\langle u, v \rangle = \langle v, u \rangle$ - Positivity: $\langle v, v \rangle \geq 0$ and $\langle v, v \rangle = 0$ iff $v = 0$ 3. **Check linearity:** Let $u, w, v \in \mathbb{R}^2$ and $a,b \in \mathbb{R}$. $$\langle au + bw, v \rangle = 3(au_1 + bw_1)v_1 + 5(au_2 + bw_2)v_2 = a(3u_1v_1 + 5u_2v_2) + b(3w_1v_1 + 5w_2v_2) = a\langle u, v \rangle + b\langle w, v \rangle$$ Linearity holds. 4. **Check symmetry:** $$\langle u, v \rangle = 3u_1v_1 + 5u_2v_2 = 3v_1u_1 + 5v_2u_2 = \langle v, u \rangle$$ Symmetry holds. 5. **Check positivity:** $$\langle v, v \rangle = 3v_1^2 + 5v_2^2$$ Since $3 > 0$ and $5 > 0$, this sum is zero only if $v_1 = 0$ and $v_2 = 0$, i.e., $v = 0$. Positivity holds. **Conclusion:** All axioms hold, so $\langle u, v \rangle = 3u_1v_1 + 5u_2v_2$ defines a valid inner product on $\mathbb{R}^2$.