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Let $A$ and $B$ be a two subspaces of a vector space $V$ such that sum $A + B$ is not the whole of $V$. Then, can we say that there must exist a non zero vector $w$, orthogonal to every vector of subspace $(A+B)$?

I am proving a theorem where I have to use this result. I need confirmation.

Thanks

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Are these inner product spaces? –  William May 22 '12 at 5:15
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To speak of othogonality you must have more than just a vector space. Do you mean that $V$ is an inner product space? If so, this is true if and only if $V$ is finite dimensional. –  Jonas Meyer May 22 '12 at 5:16
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@srijan If you talk of orthogonality, you need an inner product. –  user17762 May 22 '12 at 5:17
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What does the sum stuff have to do with the problem? –  André Nicolas May 22 '12 at 5:20
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I am sorry it should be inner product space. –  srijan May 22 '12 at 5:20

1 Answer 1

up vote 5 down vote accepted

In a finite dimensional inner product vector space, yes, because for every subspace $W$ we have $(W^{\perp})^{\perp} = W$. In particular, if $W\neq V$, then $W^{\perp}\neq\mathbf{0}$, since $\mathbf{0}^{\perp}=V$. That $W$ is a sum is immaterial.

In the infinite dimensional case, no. You can have a proper subspace whose orthogonal complement is trivial. E.g., in the vector space of all square summable real sequences, viewed as a subspace of $\mathbb{R}^{\infty}$, the span of the basis vectors $\mathbf{e}_i$ is a proper subspace that is dense, so its orthogonal complement is trivial. The same holds in any infinite dimensional Hilbert space, by taking the span of a Hilbert basis.

Added. If your space does not have an inner product, then the very concept of "orthogonality" has no meaning, so the answer is Mu.

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Thank you very much sir. –  srijan May 22 '12 at 5:31

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