Let $z=(z_1,z_2,\dots,z_n)$ and $w=(w_1,w_2,\dots,w_n)$ in $\mathbb{C}^n$. Define the inner product of $z$ and $w$ as $$\langle z,w \rangle=z_1\overline{w_1}+z_2\overline{w_2}+\dots+z_n\overline{w_n},$$ where $\overline{w_j}$ is complex conjugate of $w_j$, $j=1,2,\dots,n$.

Since $\langle z,w \rangle \neq \langle w,z \rangle$, how can we say that $z$ and $w$ are orthogonal?


1 Answer 1


We have that $$\langle x,y \rangle = \overline{\langle y,x \rangle}$$ And we call $x$ and $y$ ortogonal if $$\langle x, y \rangle = 0$$

  • $\begingroup$ Should $\langle y,x \rangle=0$ too? Or $\langle x,y \rangle=0$ if and only if $\langle y,x \rangle=0$? $\endgroup$ Jan 12, 2020 at 8:08
  • $\begingroup$ @RANGGAJAYACIPTAWAN If $\langle x,y \rangle = 0$ then $\langle y, x \rangle = \overline{\langle x,y \rangle} = \overline{0} = 0$. So we have that $x$ is orthogonal to $y$ if and only if $y$ is orthogonal to $x$, which is what we want. $\endgroup$
    – Botond
    Jan 12, 2020 at 8:11
  • $\begingroup$ Ahh i see. Thank you very much. $\endgroup$ Jan 12, 2020 at 8:13
  • $\begingroup$ @RANGGAJAYACIPTAWAN Orthogonality is symmetric $\endgroup$
    – GoodWilly
    Jan 12, 2020 at 8:42

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .