# An inner product space has a vector related to an arbitrary collection of scalars

Let $V$ be a finite dimensional inner product space with inner product $\left\langle , \right\rangle$ and let $\beta = \{v_1,...,v_n\}$ be a basis for $V$. If $c_1,...,c_n$ are arbitrary scalars, show $\exists ! \alpha \in V$ with $\left\langle \alpha , v_i \right\rangle = c_i$ $\forall i \in \{1,...,n\}$.

Here's what I've thought so far. It would be easy if $\beta$ were given as an orthogonal basis, as then $\left\langle \sum_{i=1}^n c_iv_i , v_j\right\rangle = c_i \delta_{ij}$ so $\alpha = \sum_{i=1}^n c_iv_i$ is what we require.

I know we can use the Gram-Schmidt process to acquire an orthonormal basis $\{ w_1,...,w_n\}$ from $\{v_1,...,v_n\}$ but if $\alpha = \sum_{i=1}^n c_iw_i$, there's no guarantee (that I can see) that $\left\langle w_i,v_j\right\rangle = \delta_{ij}$ to get what we require.

I was also thinking we could use the matrix $M$ for $\left\langle , \right\rangle$ to get the following system of $n$ equations; $xMv_i = c_i$ in $x$, but I don't know that this would have to have a solution.

Any ideas? Thanks!

• One thing worth thinking about is that you need only consider the cases, indexed by $j = 1, \ldots, n$, where $c_i = \delta_{ij}$. If you can find some $\alpha_j$ satisfying each of these cases, then the general case for any $c_i$ is formed by,$$\alpha = \sum_{j=1}^n c_j\alpha_j.$$ – Theo Bendit Jul 23 '15 at 18:52
• I see what you mean @TheoBendit, but I also like the hint posted already. – walkar Jul 23 '15 at 18:54

Hint: Show that $L : V \to \mathbb{R}^n$ defined by
$$L(v) = \begin{bmatrix} \langle v, v_1\rangle\\ \vdots\\ \langle v, v_n\rangle\end{bmatrix}$$