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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 $\langle w,0 \rangle = \overline{\langle 0,w \rangle} = \overline{0} = 0$.

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

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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.

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  • $\begingroup$ Thanks, yeah i realised the same thing soon after posting. $\endgroup$ Commented Apr 1, 2020 at 8:30

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