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Symmetric sesqui-linear form over $\mathbb C$ is zero

Solution to Linear Algebra Hoffman & Kunze Chapter 9.2 Exercise 9.2.4

Solution: Since $f$ is a form, we have $f$ is linear on $\alpha$. Since $f(\alpha,\beta)=f(\beta,\alpha)$, we also have $f$ is linear on $\beta$. Therefore, $f$ is a form which is also a bilinear form. Then we have\[-if(\alpha,\beta)=f(\alpha,i\beta)=f(i\beta,\alpha)=if(\beta,\alpha)=if(\alpha,\beta).\]Thus $f(\alpha,\beta)=0$. That is $f=0$.


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