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I'm considering the following problem:

Let U and V be two subspaces of $\mathbb{R}^{N}$ . True or False:

If U is orthogonal to V then U$^{\perp}$ is orthogonal to V$^{\perp}$?.

My intuition tells me that the statement is true, although I'm not sure. I know that if $\vec{v} \in V \space then \space \vec{v} \in U^{\perp}$ because the orthogonal complement $U^{\perp}$ is the set of all vectors that are orthogonal to U, and similarly if $\vec{u} \in U \space then \space \vec{u} \in V^{\perp}$.

The reason I am unsure is because there may be some vector $\vec{w} \in U^{\perp} \space and \notin V$ such that $\vec{w} \cdot \vec{v}_{1} \neq 0$ for some $\vec{v}_{1} \in V^{\perp}$, although I'm unsure how to approach proving/disproving this. Can anyone help me find a perspective from which to approach this problem?

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Your conjecture (or the conjecture made by the exercise you're working on) is false.

Counterexample:

Consider $U=\operatorname{span}(1,0,0)$ and $V=\operatorname{span}(0,1,0)$. You can confirm that $U$ is orthogonal to $V$.

We see that $U^\bot = \operatorname{span}[(0,1,0),(0,0,1)]$ and $V^\bot = \operatorname{span}[(1,0,0),(0,0,1)]$. You can confirm that $U^\bot$ is NOT orthogonal to $V^\bot$.

The general idea here is that if two subspaces are orthogonal, but they are not orthogonal complements of each other, then there must be other vectors in the space orthogonal to both. Thus their orthogonal complements will overlap a bit.

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  • $\begingroup$ Both $U^\perp$ and $V^\perp$ contain (0,0,1), so they're not linearly independent, and thus not orthogonal, correct? Thank you! $\endgroup$ – user277321 Oct 5 '15 at 23:29
  • $\begingroup$ We don't consider subspaces to be linearly independent or dependent. The way I'd state it is $(0,0,1) \in U$ and $(0,0,1)\in V$. But $(0,0,1) \cdot (0,0,1) = 1 \ne 0$. Thus these two subspaces are not orthogonal. $\endgroup$ – user137731 Oct 5 '15 at 23:31
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There is only one case for this to be true: if $\dim U+\dim V=N$, in which case $U$ and $V$ are orthogonal complements of each other.

If $\dim U=r$, $\dim V=s$ and $r+s<N$, we have $\dim U^{\bot}=N-k$, $\dim V^{\bot}=N-l$, hence $\dim U^{\bot}+\dim V^{\bot}=2N-(k+l)>N$, which implies $U^{\bot}\cap V^{\bot}\ne 0$. However, an element of $U^{\bot}\cap V^{\bot}$ can't be orthogonal to itself, unless it is $0$.

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