Exercise 6.1.22

And check those condition one by one.

• $$⟨x+z,y⟩=⟨\sum _{i=1}^n(a_i+d_i)v_i,\sum _{i=1}^n{b}_i{v}_i⟩$$

  $$=\sum _{i=1}^n(a_i+d_i)\overline{b_i}=\sum _{i=1}^n{a}_i\overline{b_i}+\sum _{i=1}^n{d}_i\overline{b_i}$$

  $$=⟨x,y⟩+⟨z,y⟩.$$

• $$⟨\mathit{cx},y⟩=⟨\sum _{i=1}^n(c{a}_i)v_i,\sum _{i=1}^n{b}_i{v}_i⟩$$

  $$=\sum _{i=1}^n(c{a}_i)\overline{b_i}=c\sum _{i=1}^n{a}_i\overline{b_i}=c⟨x,y⟩.$$

• $$\overline{⟨x,y⟩}=\overline{\sum _{i=1}^n{a}_i\overline{b_i}}$$

  $$=\sum _{i=1}^n\overline{a_i\overline{b_i}}=\sum _{i=1}^n{b}_i\overline{a_i}=⟨y,x⟩.$$

• $$⟨x,x⟩=\sum _{i=1}^n|a_i{\text{|}}^2>0$$

  if all $a_i$’s is not zero. That is, $x$ is not zero.

we have actually $a_i$ is the $i$-th entry of $x$. So the function is actually the standard inner product. Note that this exercise give us an idea that different basis will give a different inner product.