# Conjugate symmetry in normed spaces Reading the following text in a linear algebra book i fail to see how we can show $$\langle w,0 \rangle = 0$$? If we know $$\langle 0,w \rangle = 0$$ then through conjugate symmetry we know $$\overline{\langle w, 0 \rangle} = 0$$ how did they deduce from conjugate symmetry that $$\langle w,0 \rangle = 0?$$ (Here $$\overline{w}$$ is the complex conjuagte of $$w$$.)

I was thinking for any $$w$$ starting with $$\overline{w}$$ we know $$\langle \overline{w}, 0 \rangle = 0$$ and so $$\overline{\langle \overline{w}, 0 \rangle} = 0$$ but i don't that its the case $$\overline{\langle \overline{w}, 0 \rangle} = \langle w, 0 \rangle$$.

EDIT: Actually surely its just $$\langle w,0 \rangle = \overline{\langle 0,w \rangle} = \overline{0} = 0$$.

• What do you mean by $\overline {w}$? Apr 1, 2020 at 8:01

As you mentioned, $$\langle 0,w \rangle = 0$$ implies $$\overline{\langle w,0 \rangle} = 0$$ by conjugate symmetry. Taking conjugate of this last equation gives $$\langle w,0 \rangle = 0$$ as desired.
Note: for $$u,v \in V$$, we have $$\langle u,v \rangle \in \mathbb{C}$$, so that taking conjugate $$\overline{\langle u,v \rangle}$$ makes sense. However, it is not clear what does $$\overline{u}$$ mean.