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Algebra, analysis and differential equations II course of the University of Oxford

Past exam of the algebra, analysis and differential equations II course of the University of Oxford

2011

Algebra

For scalars $\alpha,\beta$ and vectors $v,w,u$, use the adjoint identity and conjugate linearity in the second argument:
$$
\begin{aligned}
\langle u,T^*(\alpha v+\beta w)\rangle
&=\langle Tu,\alpha v+\beta w\rangle\\
&=\overline\alpha\langle Tu,v\rangle+\overline\beta\langle Tu,w\rangle\\
&=\overline\alpha\langle u,T^*v\rangle+\overline\beta\langle u,T^*w\rangle\\
&=\langle u,\alpha T^*v+\beta T^*w\rangle.
\end{aligned}
$$
The difference of the two second-argument vectors is orthogonal to every $u$. Choosing $u$ to be that difference shows its squared <norm> is zero. Therefore
$$
\boxed{T^*(\alpha v+\beta w)=\alpha T^*v+\beta T^*w.}
$$
The two conjugations are why the adjoint is a linear map, despite the conjugate-linear dependence of the representing functional on its second argument.

Solution

Past exam of the algebra, analysis and differential equations II course of the University of Oxford / 2011 / Algebra / 2 / b / Solution

Past exam of the algebra, analysis and differential equations II course of the University of Oxford / 2011 / Algebra / 2 / b / Solution

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Past exam of the algebra, analysis and differential equations II course of the University of Oxford / 2011 / Algebra / 2 / b / Solution

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