UPDATE: There's an error in the question I was given.
I would appreciate it if you could help me with the following problem I have in one of my Linear Algebra course questions:
We have an inner product space $\Bbb{V}$ over $\Bbb{R}/\Bbb{C}$, and $f\in\Bbb{V}^*$ ($f:\Bbb{V}\rightarrow\Bbb{R}/\Bbb{C}$). We also have an orthonormal basis : ${v_1,...,v_n}$.
Prove that $u = \sum_{i=1}^{n}f(v_i)v_i$ is the only vector that satisfies: for every $v\in\Bbb{V}$, $f(v) = \langle v,u\rangle$.
I need help with showing that $u$ indeed satisfies the condition. Now I divided this into 2 cases: The first case is when $dim(ker(f)) = n \iff f =0$ and then $u=0$ and it works.
The second case is that $dimker(f) = n-1$ and assuming $[ v_2, ... , v_{n-1}]$ is a basis for $ker(f)$ then : $$ u = \sum_{i=1}^{n}f(v_i)v_i = f(v_1)v_1 $$ now let $v\in\Bbb{V}$, say $v = \sum_{i=1}^{n}a_iv_i$ then $$ \langle v,u\rangle = \sum_{i=1}^{n}\langle a_iv_i, f(v_1)v_1\rangle = \sum_{i=1}^{n}a_i\bar f(v_1)\langle v_i,v_1\rangle = a_1\bar f(v_1) $$ but $$ f(v) = f(\sum_{i=1}^{n}a_iv_i) = \sum_{i=1}^na_if(v_1) = a_1f(v_1) \neq a_1\bar f(v1) $$ This only works for $\Bbb {R}$. Can you tell what is the problem? Thank you.
Assume $V$ is a vector space over real. Let $v=\sum_i a_i v_i$
By linearity of $f$: $f(v) = f(\sum_i a_i v_i) = \sum_i f( a_i v_i) = \sum_i a_i f(v_i)$
$<v,u> = <\sum_i a_i v_i,\sum_i f(v_i) v_i > = \sum_{i,j} f(v_i)a_j < v_i ,v_j> = \sum_{i} f(v_i)a_i = f(\sum_i a_i v_i) = f(v)$.
In the above we used the fact: $v_1,..,v_n$ are orthonormal, we have $<v_i,v_j> = 1$ if $i=j$ and we have $<v_i,v_j> = 0$ if $i \neq j$