Chapter 5 - Inner Product Spaces - 5.1 Definition of an Inner Product Space - Problems - Page 351: 18

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Given: $v=(v_1,v_2) \in R^2$ 1. $

=2v_1v_1+v_1v_2+v_2v_1+2v_2v_2\\ =2(v_1)^2+v_1v_2+v_2v_1+2(v_2)^2\\ =(v_1)^2+(v_1)^2+2v_1v_2+(v_2)^2\\ =(v_1)^2+(v_1+v_2)^2+(v_2)^2 \geq 0$ If $v =(0,0) \rightarrow =0^2+(0+0)^2+0^2=0$ If $v \ne 0 \rightarrow =(v_1)^2+(v_1+v_2)^2+(v_2)^2 \gt 0$ if $(v_1)^2 \geq 0 ;(v_2)^2 \geq 0$ Hence, $=0$ if and only if $v=(0,0)$ 2. $=2v_1w_1+v_1w_2+v_2w_1+2v_2w_2\\ =2w_1v_1+w_2v_1+w_1v_2+2w_2v_2\\ =2w_1v_1+w_1v_2+w_2v_1+2w_2v_2\\ =$ 3. With scalar $k$, we have $kv=(kv_1,kv_2)$ and $=2(kv_1)w_1+(kv_1)w_2+(kv_2)w_1+2(kv_2)w_2\\ =2kv_1w_1+kv_1w_2+kv_2w_1+2kv_2w_2\\ =k(2v_1w_1+v_1w_2+v_2w_1+2v_2w_2)\\ =k$ 4. Let $u=(u_1,u_2)$ in $P_1(R)$ $u+v=(u_1,u_2)+(v_1,v_2)\\ =(u_1+v_1,u_2+v_2)\\ \rightarrow =2(u_1+v_1)w_1+(u_1+v_1)w_2+(u_2+v_2)w_1+2(u_2+v_2)w_2\\ =(2u_1w_1+u_1w_2+u_2w_1+2u_2w_2)+(2v_1w_1+v_1w_2+v_2w_1+3v_2w_2)\\ =+$