Exercise 6.1.5

We prove the condition for an inner product space one by one.

• $$⟨x+z,y⟩=(x+z)A{y}^{\text{∗}}=\mathit{xA}{Y}^{\text{∗}}+\mathit{zA}{y}^{\text{∗}}=⟨x,y⟩+⟨z,y⟩.$$

• $$⟨\mathit{cx},y⟩=(\mathit{cx})A{y}^{\text{∗}}=c(\mathit{xA}{y}^{\text{∗}})=c⟨x,y⟩.$$

• $$\overline{⟨x,y⟩}={(\mathit{xA}{y}^{\text{∗}})}^{\text{∗}}=y{A}^{\text{∗}}{x}^{\text{∗}}=\mathit{yA}{x}^{\text{∗}}=⟨y,x⟩.$$

  One of the equality use the fact $A=A^{\text{∗}}$.

• $$⟨x,x⟩=(x_1,x_2)A{(x_1,x_2)}^{\text{∗}}=\parallel x_1{\text{∥}}^2+i{x}_1\overline{x_2}-i{x}_2\overline{x_1}+2\parallel x_2{\text{∥}}^2$$

  $$=\parallel x_1{\text{∥}}^2+2\mathrm{Re}(i{x}_1\overline{x_2})+2\parallel x_2{\text{∥}}^2$$

  if $x_1$ or $x_2$ is not $0$. Here the function $\mathit{ℝe}(z)$ means the real part of a complex number $z$.

So it’s an inner product function. Also compute that