5
$\begingroup$

The book Linear and Geometric Algebra explains the following theorem in a way that I haven't been able to figure out:

If $\mathbf{A}$ and $\mathbf{B}$ are subspaces of a vector space $\mathbf{B}$ then the set of all combinations $\mathbf{a} + \mathbf{b}$ such that, $\mathbf{a} \in \mathbf{A}$ and $\mathbf{b} \in \mathbf{B}$ is called the span of $\mathbf{A}$ and $\mathbf{B}$, written as $\text{span}(\mathbf{A}, \mathbf{B})$.

Furthermore let $\mathbf{U}^{\bot}$ be the subspace consisting of all vectors orthogonal to a subspace $\mathbf{U}$, in the sense that $\mathbf{u} \in \mathbf{U}^{\top}$ if and only if, $\mathbf{u} \perp \mathbf{v}$ for all vectors $\mathbf{v} \in \mathbf{U}$.

I have of course been able to prove that $\mathbf{U}^{\bot}$ is indeed a subspace for all subspaces $\mathbf{U}$, if this turns out to be useful.

The theorem I want to prove is: if $\mathbf{U}$ is a subspace of $\mathbf{V}$ then $\text{span}(\mathbf{U}, \mathbf{U}^{\perp}) = \mathbf{V}$.

The book mentioned above proves it as follows: If $\text{span}(\mathbf{U}, \mathbf{U}^{\perp}) \neq \mathbf{V}$ then there is a nonzero $\mathbf{u} \perp \text{span}(\mathbf{U}, \mathbf{U}^{\perp})$ because any orthonormal basis for a subspace of an inner product space can be extended into an orthonormal basis for the entire inner product space. In particular $\mathbf{u} \perp \mathbf{U}$, i.e. $\mathbf{u} \in \mathbf{U}^{\perp}$, a contradiction.

I understand how an orthonormal basis for a subspace of an inner product space can be extended into an orthonormal basis for the whole inner product space essentially using Gram-Schmidt orthogonalisation. I don't understand how this process allows you to go from, $\text{span}(\mathbf{U}, \mathbf{U}^{\perp}) \neq \mathbf{V}$ to $\exists \mathbf{u} \in \mathbf{U} : \mathbf{u} \perp \text{span}(\mathbf{U}, \mathbf{U}^{\perp})$. So my question would be how does this implication work?

$\endgroup$
5
  • $\begingroup$ Possible duplicate of math.stackexchange.com/questions/878438/… $\endgroup$
    – Surb
    Aug 13, 2014 at 13:34
  • 1
    $\begingroup$ It seems to me that such statement holds only if $U$ is closed -- this is satisfied if everything is finite dimensional, for example.. $\endgroup$ Aug 13, 2014 at 13:35
  • $\begingroup$ Here's a counter-example for infinite dimensional spaces: math.stackexchange.com/questions/636517/… $\endgroup$
    – Surb
    Aug 13, 2014 at 13:37
  • $\begingroup$ Sure, the book I'm using is only on finite dimensional vector spaces. $\endgroup$ Aug 13, 2014 at 13:37
  • $\begingroup$ I don't think this question is a duplicate of that one because the proof in question is quite different, that one is constructive, this one is a proof by contradiction. $\endgroup$ Aug 13, 2014 at 13:41

2 Answers 2

3
$\begingroup$

You pick an orthonormal basis for $\text{span}(\mathbf{U}, \mathbf{U}^{\perp}) \neq \mathbf{V}$, say $e_j$, $1\leq j\leq n$. Extend this to an orthonormal basis for $\mathbf{V}$, $e_j$, $1\leq j\leq m$. Since $\text{span}(\mathbf{U}, \mathbf{U}^{\perp}) \neq \mathbf{V}$, $n<m$. But then $u=e_{n+1}\perp\text{span}(\mathbf{U}, \mathbf{U}^{\perp})$, by construction, and thus $u\perp \mathbf{U}$, i.e. $u\in \mathbf{U}^\perp$, and thus $u\perp u$, i.e. $\|u\|=0$. This is a contradiction, since $\|u\|=1$.

$\endgroup$
1
  • $\begingroup$ Obvious in hindsight $\endgroup$ Aug 13, 2014 at 13:50
0
$\begingroup$

Note that $Null(U)=U^\perp$ now by the rank-nullity theorem we have $Null(U)+Rank(U)=n$, where n is the dimension of $\mathbb{R}^n$. Now we can use basis extension to combine the basis for $U$ and $U^\perp$. Since $U+U^\perp$ is a subspace of $\mathbb{R}^n$, and the basis for $U+U^\perp$ has n vectors (by rank nullity thorem before), the same amount of vectors needed to span $\mathbb{R}^n$, by the $Dimension\text{ }Theorem$ the basis for $U\cup U^\perp$ is also a basis to $\mathbb{R}^n$ (It takes any set of n linearly independent vectors for them to span a subspace of dimension n). Therefore $U\cup U^\perp$=$\mathbb{R}^n$ I'm not sure how rigorous this proof is, I like the one above better as well. I just forgot this and looked up a proof, but also wanted to find my own after reading Jonas'. This idea definitely works and I thought I'd share it as an alternative.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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