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I would like to show the theorem given below:

Let $V$ be a vector space with $\dim(V)=n$, let $U$ be a real subspace of $V$ (i.e. $U\subset V$ and $U\neq 0$) and let $v\in V\setminus U$. Then there exists a $f\in V^*$ s.t. $f(u)=0$ for all $u\in U$ and $f(v)=1$.

I started with taking a basis $B_u=(u_1,\cdots, u_k), \,(k<n)$ for $U$, which can be extended to a basis $B_v=(u_1,\cdots,u_k,u_{k+1},\cdots,u_n)$ for $V$. Then there exists a dual basis $B^*=(v_1^*,\cdots,v_n^*)$ s.t. $v_i^*(u_j)=\delta_{i,j}$.

Using these definitions I started to compute $f(u)$ for an arbitrary $u\in U$:

$f(u)=f(\sum\limits_{i=1}^{k}\lambda_i u_i)=\sum\limits_{i=1}^k\lambda_i f(u_i)=\sum\limits_{i=1}^k \lambda_i (\sum\limits_{j=1}^nf(u_j)v_j^*(u_i))$

At this point I am struggling to find the next steps. It kind of looks like the result can be seen by considering the $v_j^*(u_i)$ parts because they either evaluate to $1$ or $0$, but I am not sure how to argue that this term is $0$ and, if I consider $v$, i.e. extend the outer sum for $v=\sum\limits_{i=1}^n \mu_i u_i$, that the term is $1$.

I would be very grateful, if someone can help me.

Sebastiano
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NeverEver
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  • $f(x)=\frac{\langle v, x\rangle}{\langle v,v\rangle}$ does it, since $v\notin U \implies v\in U^\perp$. I must add, any finite dimensional vector space (over $\mathbb{C}$ or $\mathbb{R}$ at least) can be made into an inner product space, check https://math.stackexchange.com/questions/352120/inner-product-for-a-finite-dimensional-vector-space?rq=1. – theGrey Nov 08 '22 at 18:23
  • This is a good idea. I have one question for clarification: We can assume $v\neq 0$ because $U$ is a subspace, which always contains the zero vector, right? – NeverEver Nov 08 '22 at 19:53
  • yes, also because if $v$ is the zero vector and $g:V\to X$ is any linear function between vector spaces then $f(v)=0_X$ – theGrey Nov 08 '22 at 21:48
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    @theGrey - $v \notin U \not\implies v \in U^\perp$. Indeed, for any non-zero $u \in U, u^\perp \in U^\perp, a \in \Bbb R, b \in \Bbb R$, if $v = au + bu^\perp$, then $v$ will not be in $U$ or $U^\perp$. – Paul Sinclair Nov 09 '22 at 16:14
  • @PaulSinclair you are right I was sloppy. Then take the same $f$ as in my first comment with $v$ replaced by $v-P_U(v)$ i.e. $v$ is replaced by ita orthogonal component wrt to $U$ – theGrey Nov 09 '22 at 17:45
  • I suggest extending your basis for U by first adding $v$, then other vectors. With $v$ as a basis vector, you can write any other vector in terms of this basis. Now consider the coefficient of $v$ in that expansion. – Paul Sinclair Nov 09 '22 at 18:30

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