1
$\begingroup$

For a non-zero vector $\mathbf{v} \in{\mathbb{E}^n}$, I need to show that the collection $W$ of vectors orthogonal to $\mathbf{v}$ forms an (n-1)-dimensional subspace of $\mathbb{E}^n$. I've been working with a spanning set $\alpha_1\mathbf{w}_1 + \dotsb + \alpha_{n-1}\mathbf{w}_{n-1}$, but I'm having trouble trying to wrap my head around how to prove this is linearly independent or why it has to have dimension of $(n-1)$. Thanks

$\endgroup$
  • 1
    $\begingroup$ Do you have the Gram-Schmidt process available? $\endgroup$ – Henning Makholm Oct 11 '11 at 0:00
  • 1
    $\begingroup$ Here's one way of approaching this: suppose $\{w_1, \ldots, w_k \}$ is a basis for $W$. Show that $\{w_1, \ldots , w_k, v\}$ is a basis for $\mathbb{E}^n$. Since every basis of $\mathbb{E}^n$ has $n$ elements, we have $k+1=n$. $\endgroup$ – Chris Eagle Oct 11 '11 at 0:02
  • $\begingroup$ Unfortunately not, I've read about it but apparently this is solvable without using it... $\endgroup$ – James H Oct 11 '11 at 0:04
  • $\begingroup$ At least, then, skimming through Wikipedia's account of Gram-Schmidt ought to inspire an attack on the missing piece of Chris's hint. $\endgroup$ – Henning Makholm Oct 11 '11 at 0:25
2
$\begingroup$

Since $v$ is non-zero, then the set $\{v\}$, which is linearly independent, can be complete to a basis $B=\{v,w_1,\dots,w_n\}$ of $E^n$, which necessarily must have $n$ elements. Now consider the vectors $$w_i'=w_i-\frac{v\cdot w_i}{v\cdot v}v$$ for $i\in\{1,\dots,n\}$. An easy computation shows that the $w_i'$ are orthogonal to $v$. You should next check that they are linearly independent.

$\endgroup$
2
$\begingroup$

I think the simplest way is this. Let $${\bf v}=(a_1,a_2,\dots,a_n), {\bf w}=(w_1,w_2,\dots,w_n)$$ The orthogonality condition is a single homnogeneous linear equation in $n$ unknowns, $$a_1w_1+a_2w_2+\cdots+a_nw_n=0$$ Do you know how to find the dimension of the solution space of a system of homogeneous equations?

EDIT: if not, then see my comment, currently the third one down from this answer.

(Aside to Agusti Roig: perhaps I should have done as you suggested, but what I've done instead should still help, as well as saving a lot of typing!)

$\endgroup$
  • $\begingroup$ no but this sounds very helpful $\endgroup$ – James H Oct 11 '11 at 1:11
  • $\begingroup$ Hint: apply the rank-nullity theorem. :-) $\endgroup$ – d.t. Oct 11 '11 at 1:38
  • 2
    $\begingroup$ Or, if you don't know the rank-nullity theorem, do this: at least one of the $a_i$ is not zero. Let's assume $a_n\ne0$. Then one solution has $w_1=1$, $w_n=-a_1/a_n$, and all the other $w_i=0$. A second solution has $w_2=1$, $w_n=-a_2/a_n$, all the other $w_i=0$. And so on, up to solution number $n-1$, which has $w_{n-1}=1$, $w_n=-a_{n-1}/a_n$, all the other $w_i=0$. So we get $n-1$ elements of $W$. It's easy to check that they are linearly independent, and it isn't too hard to show that every element of $W$ is a linear combination of these elements. $\endgroup$ – Gerry Myerson Oct 11 '11 at 2:34
  • $\begingroup$ @Gerry. You're right: your idea is the simplest one. But maybe you could add this comment of yours in your answer: perhaps it would help james. $\endgroup$ – d.t. Oct 11 '11 at 8:36
1
$\begingroup$

You can do it in several ways.

First, you can consider a map

$$ f: \mathbb{E}^n \longrightarrow \mathbb{R} \ , \qquad f(u) = v\cdot u \ . $$

Prove that $f$ is linear. By the rank-nullity theorem, you'll have

$$ \mathrm{dim}\, \mathbb{E}^n = \mathrm{dim}\,\mathrm{Im}\, f + \mathrm{dim}\,\mathrm{Ker}\, f \ . $$

And now you must prove two more things: (1) $\mathrm{Im}\, f = \mathbb{R}$ (hint: consider vectors $u$ of the form $\lambda v$). And (2) $\mathrm{Ker}\, f = W$. As a consequence, you'll have all: that $W$ is a subspace and its dimension.

Another approach: if you know what an (internal) direct sum is, prove that

$$ \mathbb{E}^n = \mathrm{span} (v) \oplus W \ . $$

For this, you'll need to: (1) Prove that $W$ is a subspace (hint: just look at the definition of subspace). (2) Prove that $\mathrm{span} (v) \cap W = \left\{ 0\right\} $. And (3), the tough part of this, prove that you can write every vector $u \in \mathbb{E}^n$ as $u = \lambda v + w$, for some $\lambda \in \mathbb{R}$ and $w\in W$. This can be done easily if you know about orthogonal projections. In this case, $\lambda v$ is just the orthogonal projection of $u$ onto $\mathrm{span} (v)$ and $w = u -\lambda v$. (Hint: $\lambda = \dfrac{v\cdot u}{v\cdot v}$.)

As a consequence, $\mathrm{dim}\, \mathbb{E}^n = \mathrm{dim}\,\mathrm{span} (v) + \mathrm{dim}\, W$ and you're done.

$\endgroup$
0
$\begingroup$

Suppose: A = {x$_1$, x$_2$,...,x$_{n-1}$, v} is an orthogonal basis for R$^n$ (this assumption is fine because you can just take n-1 vectors that are orthogonal to each other and v to form a basis for R$^n$). That means that any vector in R$^n$ can be expressed as a linear combination of the vectors in A. Therefore, any vector orthogonal to v in R$^n$ can be expressed as a$_1$x$_1$+ a$_2$x$_2$+...+a$_{n-1}$x$_{n-1}$+ 0 v. This means that {x$_1$, x$_2$,...,x$_{n-1}$} is a basis for all vectors orthogonal to v in R$^n$. That's why the orthogonal space has dimension n-1. After that it's just a matter of closing this set of vectors under scalar multiplication and vector addition, which isn't too bad.

$\endgroup$
  • $\begingroup$ subscripts didn't work for me i guess $\endgroup$ – Benji Huynh Jul 27 '17 at 18:17
  • 1
    $\begingroup$ use dollar sign to encapsulate mathjax syntax $\endgroup$ – Furrane Jul 27 '17 at 18:33

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.