Mathematics Asked by Ivan Bravo on December 6, 2020

I was doing an exercise and I saw this property, I would like to know why it’s true.

Let $V$ be a $mathbb{K}$-dimensional vector space and Let $f$ be a bilinear form. Given $S subseteq V$, we define: $S^perp$={$alpha in V | f(alpha, beta)=0$

$forall beta in S$}

{0}$^perp$=$V$

- By definition, $S^perp$ is subset of $V$, so ${0}^perp subseteq V$.
- On the other hand, if $alpha in V$ is arbitrary, as $v mapsto f(alpha,v)$ is linear, we have that $f(alpha,0) = 0$, meaning that $f(alpha,beta) = 0$ for every $beta in {0}$. Thus $alpha in {0}^perp$ and this shows $V subseteq {0}^perp$.

Answered by azif00 on December 6, 2020

Every $xin V$ satisfies that $f(x,0)=0$ for $f$ is bilinear. And given that ${0}$ is only composed by $0$ (obviously) then every $xin V$ is in the ortogonal complement of $0$.

Answered by Iesus Dave Sanz on December 6, 2020

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