So if V1,....Vn is a basis for R^n and S = spanV1,...Vk for some positive interger kVk+1,....Vn.
I am pretty sure that this statement is true... because S transpose cannot be greater than S or else it'll be in R^(n+1)... I am not really sure how to prove it though... I could be wrong as well :( Please help!
Actual Question: https://gyazo.com/0bdca978f9440383bd13e1a260199fda
The formulation you stated doesn't make sense, fortunately I could read your link. The statement is definetely true that \begin{equation} S^{\bot}=span(v_{k+1},..,v_{n}) \end{equation} Let say $w$ is orthogonal to $S$ which means $w\in S^{\bot}$ or in other form \begin{equation} (w,h)=0,\quad\forall h\in S \end{equation} Then since $v_{1},..v_{n}$ is a base then $w=\sum\xi^{i}v_{i}$ e \begin{equation} (w,h)=(\sum\xi^{i}v_{i},h)=\sum\xi^{i}(v_{i},h)=0 \end{equation} choosing $h=v_{1}$ this imply that $\xi^{1}=0$. Proceeding the same way if we choose $h=v_{i}$ with $i=1...k$ then we have that $\xi^{i}=0$ for $i=1...k$ which means that \begin{equation} w\in span(v_{k+1},..,v_{n}) \end{equation} To be really strict we only demonstrate that the orthogonal space include the $span(v_{k+1},..,v_{n})$, the converse is given directly confirming that $v_{k+1},..,v_{n}\in S^{\bot}$