Complex conjugate in inner products

When we solve for inner product of $$\rvert a \rangle \cdot \rvert b \rangle$$ we solve for $$\langle a \rvert b \rangle$$ where $$\langle a \rvert$$ is complex conjugate of $$\rvert a \rangle$$. However this confuses me because in linear algebra, $$u \cdot v$$ is $$uv^*$$. The latter vector is conjugated. Why does braket notation conjugate prior vector and linear algebra conjugate latter vector?

• Which variable is conjugated? Mathematics and physics use the opposite conventions for that. Your "bra" and "ket" notation $\langle a \rvert$ is only used in physics, so you probably should use that convention. – GEdgar May 29 at 11:31

In linear algebra, $$u\cdot v$$ is $$u^* v$$, the first vector gets conjugated.