Let \(V\) be a finite-dimensional vector space and let \(\mathfrak{B} = \{\alpha_1, \ldots, \alpha_n\}\) be a basis for \(V\). Let \((\quad | \quad)\) be an inner product on \(V\). If \(c_1, \ldots, c_n\) are any \(n\) scalars, show that there is exactly one vector \(\alpha\) in \(V\) such that \((\alpha \mid \alpha_j) = c_j\), \(j = 1, \ldots, n\).
Solution
Define \[ \Phi:V\longrightarrow F^n, \qquad \Phi(\alpha) =\bigl((\alpha\mid\alpha_1),\ldots,(\alpha\mid\alpha_n)\bigr). \] With the convention used in Hoffman–Kunze, the inner product is linear in its first argument, so \(\Phi\) is linear [@HK71].
We claim that \(\Phi\) is injective. If \(\Phi(\alpha)=0\), then \[ (\alpha\mid\alpha_j)=0 \] for every basis vector \(\alpha_j\). By linearity in the second slot up to the usual conjugation convention, this implies \[ (\alpha\mid\beta)=0 \] for every \(\beta\in V\), because the \(\alpha_j\) span \(V\). Taking \(\beta=\alpha\) gives \[ (\alpha\mid\alpha)=0, \] and positive definiteness forces \(\alpha=0\). Thus \(\ker\Phi=0\).
Now \[ \dim_FV=n=\dim_FF^n. \] An injective linear map between finite-dimensional spaces of the same dimension is bijective. Therefore for every \[ (c_1,\ldots,c_n)\in F^n \] there exists a unique \(\alpha\in V\) such that \[ \Phi(\alpha)=(c_1,\ldots,c_n), \] that is, \[ \boxed{(\alpha\mid\alpha_j)=c_j\quad(j=1,\ldots,n).} \]