0
$\begingroup$

Problem: Let $V$ be an inner product space, $S$ and $S_0$ be subsets of $V$, and $W$ be subsets of $V$, and $W$ be a finite-dimensional subspace of $V$. Prove 1). $S_0 \subseteq S$ implies that $S^{\bot} \subseteq S_0^{\bot}$

Brief sketch: Let $\beta={v_1,...v_k}$ be a basis for $S_0$, $\beta'{v_1',..,v_k'}$ bais for $S_0^{\bot}$, $\gamma={w_1,...,w_k}$ basi for $S$ and $\gamma'={w_1',...,w_k'}$ basis for $S^{\bot}$.

by the theorem, we can conclude the statement.

Theorem: suppose $S={v_1,v_2,..,v_k}$ is an orthonormal set in an $n$-dimensional inner product space $V$, then $S$ can be extended to an orthonormal basis ${v_1,v_2,...,v_k,v_{k+1},...,v_n}$ for $V$.

Will that be enough?

$\endgroup$
2
$\begingroup$

No, it will not be enough. What does “Let $\beta={v_1,...v_k}$ for $S_0$, $\beta'{v_1',..,v_k'}$ for $S_0^{\bot}$mean? I have no idea.

Besides, you don't need bases here. If $v\in S^\top$, then, for each $w\in S$, $\langle v,w\rangle=0$. In particular (since $S_0\subset S$), for each $w\in S$, $\langle v,w\rangle=0$, and so $v\in S_0^\top$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.