Theorems on Inner Product Space: Part-1 #
Theorems-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 \(\pmb{ L\{x_{1},x_{2},…,x_{k},y \}= L\{x_{1},x_{2},…,x_{k},x \}}\) .