# theorem of existence of an orthogonal complement [duplicate]

This question already has an answer here:

are there any theorems that give us any conditions to know if a linear subspace $E$ plus its orthogonal complement span the whole vector space?

For exemple, I know that in $\mathbb{R}[X]$, the complement orthogonal of the hyperspace $Span(1+X, 1+X^2, ...)$ is $\{0\}$ and thus the sum does not span the whole space

Thanks !

## marked as duplicate by Adrian Keister, Theoretical Economist, Namaste linear-algebra StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Sep 7 '18 at 0:42

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

• In order to define "orthogonal complement" you need an inner product, or at least a bilinear form. Which bilinear form are you using on $\mathbb R[X]$? – Robert Israel Sep 6 '18 at 20:38

## 1 Answer

Hint: you can prove that $(A + B)^T=A^T \cap B^T$, this is rather easy. Then it is sufficient to notice that $(A^T+A)^T=A^T \cap A=0$.