Theorem-1

 Let \(\pmb{ V}\) be inner product space over a field \(\pmb{ F}\) and \(\pmb{S=\{x_{1},x_{2},…,x_{k} \}}\) be a orthogonal subset of non null vectors of \(\pmb{ V}\). If \(\pmb{ x\in V-L(S)}\) and \(\pmb{y=x- \displaystyle\sum_{i=1}^{k} c_{i}x_{i}}\) then
 (a)   \(\pmb{ x}\) is orthogonal to each \(\pmb{ x_{i}}\)

 (b)   \(\pmb{ L\{x_{1},x_{2},…,x_{k},x \}=L\{x_{1},x_{2},…,x_{k},y \}}\) where \(\pmb{ c_{i}=\frac{\lang x,x_{i}\rang}{\lang x_{i},x_{i}\rang}}\), \(\pmb{ i=1,2,…,k}\) .

 Proof:   Given that \(\pmb{S=\{x_{1},x_{2},…,x_{k} \}}\) be a orthogonal subset of non null vectors of the inner product space \(\pmb{ V}\) over the field \(\pmb{ F}\).
 Let \(\pmb{ x\in V-L(S)}\) and \begin{align} \pmb{y=x- \displaystyle\sum_{i=1}^{k} c_{i}x_{i}} \end{align}   where \(\pmb{ c_{i}=\frac{\lang x,x_{i}\rang}{\lang x_{i},x_{i}\rang}}\), \(\pmb{ i=1,2,…,k}\) .
 To prove \( \pmb{ \lang y,x_{j} \rang=0,~j=1,2,…k}\).  Now for all \(\pmb{j}\), \begin{align*} \pmb{ \lang y,x_{j} \rang}& = \pmb{ \left\lang x- \displaystyle\sum_{i=1}^{k} c_{i}x_{i}~,~x_{j} \right\rang}\\ & = \pmb{ \left\lang x,~x_{j} \right\rang} – \pmb{ \left\lang \displaystyle\sum_{i=1}^{k} c_{i}x_{i}~,~x_{j} \right\rang}\\ & = \pmb{ \left\lang x,~x_{j} \right\rang} – \pmb{ \displaystyle\sum_{i=1}^{k} c_{i} \left\lang x_{i}~,~x_{j} \right\rang}\\ & = \pmb{ \left\lang x,~x_{j} \right\rang} – \pmb{c_{j} \left\lang x_{j}~,~x_{j} \right\rang~since~\lang x_{i},x_{j}\rang=0~when~i\ne j}\\ & = \pmb{ \left\lang x,~x_{j} \right\rang} – \pmb{\frac{\lang x,x_{j}\rang}{\cancel{\lang x_{j},x_{j}\rang}} \cancel{\left\lang x_{j}~,~x_{j} \right\rang}~since~\lang x_{j},x_{j}\rang\ne 0, x_{j}~ is ~non~null }\\ & = \pmb{ \cancel{\lang x,~x_{j} \rang} – \cancel{\lang x,~x_{j} \rang} }\\ & = \pmb{0} \end{align*}  Therefore \( \pmb{ \lang y,x_{j} \rang=0,~j=1,2,…k}\) .

 To prove \(\pmb{ L\{x_{1},x_{2},…,x_{k},x \}=L\{x_{1},x_{2},…,x_{k},y \}}\)

 First we prove that \(\pmb{ L\{x_{1},x_{2},…,x_{k},x \}\subset L\{x_{1},x_{2},…,x_{k},y \}}\)
 Let \begin{align*} & \pmb{z\in L\{x_{1},x_{2},…,x_{k},x \}}\\ \implies & \pmb{z=\displaystyle\sum_{i=1}^{k}p_{i}x_{i}+px~for~some~p_{i},p\in F}\\ \implies & \pmb{z=\displaystyle\sum_{i=1}^{k}p_{i}x_{i}+p \left(y+\displaystyle\sum_{i=1}^{k} c_{i}x_{i} \right)~using~(1) }\\ \implies & \pmb{z=\displaystyle\sum_{i=1}^{k}p_{i}x_{i}+py+ \left(\displaystyle\sum_{i=1}^{k} pc_{i}~x_{i} \right)}\\ \implies & \pmb{z=\displaystyle\sum_{i=1}^{k}\left(p_{i}+pc_{i}\right)x_{i}+py}\\ \implies& \pmb{z\in L\{x_{1},x_{2},…,x_{k},y \}}\\ \end{align*}  Therefore \(\pmb{ L\{x_{1},x_{2},…,x_{k},x \}=L\{x_{1},x_{2},…,x_{k},y \}}\).

 Lastly we prove that \(\pmb{ L\{x_{1},x_{2},…,x_{k},y \}\subset L\{x_{1},x_{2},…,x_{k},x \}}\)
 Let \begin{align*} & \pmb{z\in L\{x_{1},x_{2},…,x_{k},y \}}\\ \implies & \pmb{z=\displaystyle\sum_{i=1}^{k}q_{i}x_{i}+qy~for~some~q_{i},q\in F}\\ \implies & \pmb{z=\displaystyle\sum_{i=1}^{k}q_{i}x_{i}+q \left(x-\displaystyle\sum_{i=1}^{k} c_{i}x_{i} \right)~using~(1) }\\ \implies & \pmb{z=\displaystyle\sum_{i=1}^{k}q_{i}x_{i}+qx- \left(\displaystyle\sum_{i=1}^{k} qc_{i}~x_{i} \right)}\\ \implies &\pmb{z=\displaystyle\sum_{i=1}^{k}\left(q_{i}-qc_{i}\right)x_{i}+qx}\\ \implies& \pmb{z\in L\{x_{1},x_{2},…,x_{k},x \}}\\ \end{align*}  Therefore \(\pmb{ L\{x_{1},x_{2},…,x_{k},y \}\subset L\{x_{1},x_{2},…,x_{k},x \}}\) .
 Hence \(\pmb{ L\{x_{1},x_{2},…,x_{k},y \}= L\{x_{1},x_{2},…,x_{k},x \}}\) .