Orthogonal decomposition using Gram determinants

Theorem. Let $\mathcal{H}$ be a Hilbert space equipped with the inner product $\langle \cdot, \cdot \rangle$, and let $w, v_1, \ldots, v_n \in \mathcal{H}$. If $$G(w_1, \ldots, w_n) = \det [\langle w_i, w_j \rangle]_{i,j=1}^{n}$$ denotes the Gram determinant, then the followings hold true:

1. The vector $w_{\perp}$ defined by the (formal) determinant $$w_{\perp} = \frac{1}{G(v_1,\ldots,v_n)}\begin{vmatrix} w & v_1 & \cdots & v_n \\ \langle v_1, w \rangle & \langle v_1, v_1 \rangle & \cdots & \langle v_1, v_n \rangle \\ \vdots & \vdots & \ddots & \vdots \\ \langle v_n, w \rangle & \langle v_n, v_1 \rangle & \cdots & \langle v_n, v_n \rangle \end{vmatrix} \tag{1}$$ is orthogonal to $V = \operatorname{span}\{v_1, \ldots, v_n\}$, and in fact, $$w = w_{\perp} + (w - w_{\perp})$$ is the orthogonal decomposition of $w$.
2. The square-distance between $w$ and $V$ is given by $$\|w_{\perp}\|^2 = \operatorname{dist}(w, V)^2 = \frac{G(w, v_1, \ldots, v_n)}{G(v_1, \ldots, v_n)}. \tag{2}$$

Proof. Extend the notation by letting $$G(\{v_i\}_{i=1}^{n}, \{w_i\}_{i=1}^{n}) = \det [\langle v_i, w_j \rangle]_{i,j=1}^{n},$$ and then define the linear functional $\ell$ on $\mathcal{H}$ by $$\ell(u) = G(\{u, v_1, \ldots, v_n\}, \{w, v_1, \ldots, v_n\}).$$ By the Riesz representation theorem, there exists $h \in \mathcal{H}$ such that $\ell(u) = \langle h, u \rangle$ for all $u \in \mathcal{H}$. Since $\ell(v_i) = 0$ for all $i = 1, \ldots, n$, it follows that $h$ is orthogonal to $V$. Moreover, expanding the determinant defining $\ell(u)$ along the first line, we see that $$\begin{align*} \ell(u) &= G(v_1, \ldots, v_n) \langle u, w \rangle \\ &\quad + \sum_{i=1}^{n} (-1)^i G(\{w, v_1, \ldots, \widehat{v_i}, \ldots, v_n\}, \{v_1, \ldots, v_n\}) \langle u, v_i \rangle, \end{align*}$$ and so, $$\begin{align*} h &= G(v_1, \ldots, v_n) w \\ &\quad + \sum_{i=1}^{n} (-1)^i G(\{w, v_1, \ldots, \widehat{v_i}, \ldots, v_n\}, \{v_1, \ldots, v_n\}) v_1, \end{align*}$$ which is precisely the formal determant in $\text{(1)}$. This also implies that $h$ is a linear combination of $w, v_1, \ldots, v_n$ with the coefficient of $w$ given by $G(v_1, \ldots, v_n)$, hence $$w_{\perp} = \frac{h}{G(v_1, \ldots, v_n)} \in w + V.$$ This and $w_{\perp} \perp V$ together proves the first item of the theorem. For the second item, the equality $\|w_{\perp}\| = \operatorname{dist}(w, V)$ is obvious by the orthogonality. Moreover, $$\|w_{\perp}\|^2 = \langle w_{\perp}, w_{\perp} \rangle = \langle w_{\perp}, w \rangle = \frac{\ell(w)}{G(v_1, \ldots, v_n)} = \frac{G(w, v_1, \ldots, v_n)}{G(v_1, \ldots, v_n)}.$$ This completes the proof. $\square$