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I am also given that if $S$ and $T$ are subspaces of a vector field, then the above are equivalent

Martin Sleziak
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user58503
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2 Answers2

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Suppose $u \perp S$ and $v \perp T$; that is, $\langle u,s \rangle = 0$ and $\langle v,t \rangle = 0$ for all $s \in S$ and $t \in T$. You need to show that $u+v \perp S \cap T$; that is, $\langle u+v, x \rangle = 0$ for all $x \in S \cap T$. Can you do this?

Now suppose $S$ and $T$ are subspaces. Why does the converse to the above hold?

If you're still stuck, please post some of your working.

Clive Newstead
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  • My problem is this,

    If I have that (S + T) perp is a subset of (S perp intersection T perp), can I conclude that:

    (S perp intersection T perp) perp is a subset of ( S + T )?

    If so, why? i.e. by which rule?

     – user58503 Jan 16 '13 at 17:52
  • Yes - general if $A \subseteq B$ then $B^{\perp} \subseteq A^{\perp}$; and if $A$ is a subspace then $A^{\perp \perp} = A$. To see why this is true, just work from the definitions, in a similar way to how I did in my answer. – Clive Newstead Jan 16 '13 at 19:00
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If $x\in S^\perp+T^\perp$, then we can write $x=u+v$ with $u\in S^\perp$ and $v\in T^\perp$. Now if $r\in S\cap T$, then $u\perp r$ and $v\perp r$, hence $x=u+v\perp r$, i.e. $x\in(S\cap T)^\perp$.

Hagen von Eitzen
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