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It's a problem in a textbook, offering no answers. I have try my best to solve it, but no result.

For $V$ is a linear closure in $\mathbb{R}^n$. If $V=U\oplus W$ is a direct sum decomposition , then call $W$ the complement of $U$ in $V$. $U$ is for $W$ the same. Is the complement of $U$ in $V$ is uniquely determined? And compare $W$ with the complement $V\setminus U$ under the concept of set theory.

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    What is a linear closure? Is it just a subspace?2012-11-19

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The complement is not uniquely determined (although it's dimension is). Consider the complement of the subspace $U$ spanned by $\begin{pmatrix}1 & 0\end{pmatrix}^\mathrm{T}$ in $\mathbb{R}^2$. Any subspace spanned by any vector which is not a multiple of $\begin{pmatrix}1 & 0\end{pmatrix}^\mathrm{T}$ will serve as a valid complement.

This complement differs from the set theoretic complement in a major way. $V\setminus U$ is a set but not a vector space in general; it simply removes the portion of $W$ in $V$. In the above example, $\mathbb{R}^2\setminus U$ is simply the $xy$-plane with the $x$-axis carved out. It does not have the closed structure of a vector space.

A more natural analogue of the set-theoretic complement is given by $V/U$ which is a quotient space. In this structure, we not only remove the vectors of $U$ but also compress the components of the remaining vectors in the direction of $U$ to retain a vector space structure.