Let $V$ a vector space with $dim(V)<\infty$. Let $v^*\in V^*\setminus\{0\}$, show that there exists a $v\in V$ with $v^*(v)=1$.$\quad (V^*$ denotes the dual space.)
My approach: Let $B=(v_1,\dots,v_n)$ a basis of of V and $B^*=(v_1^*,\dots,v_n^*)$ its corresponding dual basis . Let $v\in V$ arbitrary. Then we can find a linear combination of $\lambda_i,v_i$ such that $v=\sum_{i=1}^n\lambda_iv_i$. Furthermore we can find $\mu_i,v_i^*$ such that we can express $v^*\ne0 $ with $v^*=\sum_{i=1}^n \mu_iv_i^*$. Also $v^*$ is a linear function. Thus we can conclude since $v_i^*(v_j)=\delta_{i,j}$ $$v^*(v)=v^*\Bigl(\sum_{i=1}^n\lambda_iv_i \Bigr)=\sum_{i=1}^n \lambda_i v^*(v_i)=\sum_{i=1}^n\lambda_i \sum_{j=1}^n\mu_jv_j^*(v_i)=\lambda_i\mu_i$$
I do not see how to proceed here. This attempt would hold if we can take $\lambda_i^{-1}=\mu_i$, but why should we be allowed to do so?
Some help would be graceful!
The fact that $v^* \in V^* \setminus \{0\}$, i.e., $v^* \neq 0$, tells us that there is some $v_0 \in V$ such that $v^*(v_0) \neq 0$. Thus we can define $$v = \frac{v_0}{v^*(v_0)},$$ and then $$v^*(v) = v^*\left(\frac{v_0}{v^*(v_0)}\right) = \frac{1}{v^*(v_0)}v^*(v_0) = 1.$$ In particular, the condition $\mathrm{dim}(V) < \infty$ is not necessary.
To conclude your own solution: first note that you are missing a summation sign in the final term of your displayed equation. It should be $\sum_i \lambda_i\mu_i$. Now note that at least one of the $\mu_i$ must be non-zero; otherwise $v^* = 0$. Hence you can set the corresponding $\lambda_i$ to $1/\mu_i$, and set all the other $\lambda_i$ to 0.