Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
    
What is a linear closure? Is it just a subspace? –  EuYu Nov 19 '12 at 16:05

1 Answer 1

up vote 0 down vote accepted

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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.