Complex conjugate in inner products

Question

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?

Answer 1

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