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Let $U$ and $W$ be subspaces of an inner product space $V$. If $U$ is a subspace of $W$, then $W^{\bot}$ is a subspace of $U^{\bot}$?.

I don't find the above statement intuitively obvious. Could someone provide a proof?

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2 Answers 2

up vote 4 down vote accepted

If you're orthogonal to everything in a set, then you're also orthogonal to everything in every subset of that set.

Put another way: Elements of $W^\perp$ have to be orthogonal to more vectors than elements of $U^\perp$; they have to be orthogonal to the vectors in $U$ and the vectors in $W\setminus U$ (if any). Therefore there are less of them than if we took the vectors that only have to satisfy the property of being orthogonal to vectors in $U$. (More restrictions$\implies$ Fewer vectors satisfying the restrictions.)

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And that since you know that $U^\perp,W^\perp$ are subspaces (hopefully) of $V$ then it suffices to show containment. –  Alex Youcis Nov 6 '11 at 22:30
    
So shouldn't it be $U^{\bot}$ is a subspace of $W^{\bot}$ ? That is my confusion. –  Mark Nov 6 '11 at 22:33
2  
@Mark: No. If you're orthogonal to everything in $W$, i.e., if you're in $W^\perp$, then you're also orthogonal to everything in a subset of $W$ such as $U$, i.e., you're in $U^\perp$. For a concrete example, think of $\mathbb R^3$ with standard dot product, let $U$ be the span of $(1,0,0)$, and let $W$ be the span of $\{(1,0,0),(0,1,0)\}$. If you have to be orthogonal to both $(1,0,0)$ and $(0,1,0)$, then you are in particular orthogonal to $(1,0,0)$, so $W^\perp\subset U^\perp$. –  Jonas Meyer Nov 6 '11 at 22:35
    
Oh right, I had this at the back of my head, but was unable to explain it clearly. Now I understand, thanks. –  Mark Nov 6 '11 at 22:54

It should be intuitive, already at the level of logic:

To be in $W^\perp$, you have to satisfy a certain condition $P(w)$ (namely: 'be orthogonal to $w$') for each and every element $w\in W$.

So given a subset $U\subseteq W$, to be in $U^\perp$ means you have to satisfy $P(w)$ merely for all $u\in U$.

Thus you have to satisfy less properties to be in $U^\perp$, thus it is easier to be in $U^\perp$, thus $U^\perp$ is larger: $W^\perp\subseteq U^\perp$.

It should also be intuitive geometrically: consider $\mathbb{R}^3$, let $U$ be the $x$-axis, and $W$ the $x,y$-plane. Then $U^\perp$ is the $y,z$-plane, and $W^\perp$ is the $z$-axis.

//Edit: I was slow so I missed Joans Meyer's edit, which kind of makes my answer redundant.

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