0
$\begingroup$

Let $V$ be a vector space of dimension $n$ and some set $E = \{v_1, v_2, \dotsc, v_n\}$. If, for every $v \in V$, we have $\begin{cases} \langle v, v_i \rangle = 0 \\ i = 1,2,3,\dotsc,n \end{cases} \implies v=0$, what can be said about $E$?

  1. Can $E$ be a basis for $V$?
  2. Can $E$ be a orthonormal basis for $V$?

What I noticed is that, for a given orthonormal basis $B = \{b_1, b_2, \dotsc, b_n\}$ for $V$, we can write $$ v = \langle b_1, v \rangle b_1 + \langle b_2, v \rangle b_2 + \dotsb + \langle b_n, v \rangle b_n $$ In this particular question, there is a problem: if $B$ is such a basis, then we can write $v$ the way I posted. Nothing is said about the opposite. I'm stuck! Could anyone, please, help?

EDIT: I fixed the question. Sorry about that.

$\endgroup$
1
$\begingroup$

Assuming the operation is an inner product, if $v$ is in the span of $E$, write $v = \sum a_i v_i$. Then either $v = 0$, or $0 < \langle v, v \rangle = \langle \sum a_iv_i, v\rangle = \sum (a_i\langle v_i, v \rangle) = 0$, which is a contradiction.

EDIT: Ok, I see your edit. To answer your questions as asked: yes, $E$ can be an orthonormal basis. Take the one dimensional vector space $\mathbb R$ and let $E = \{ v_1 \}$ contain any nonzero vector.

To answer the question I think you might have meant to ask: your condition does not imply that $E$ is a basis. Add the zero vector into $E$ in the previous example.

However, suppose we take a maximal linearly independent subset $E' \subseteq E$. This set is nonempty (otherwise $E$ would contain either nothing or just the zero vector, and not satisfy your condition). Let $W$ be the subspace spanned by $E'$. We can show that $V = W \oplus W^\perp$ (if you're not familiar with this theorem, it's very important and would make a good followup question). If $W^\perp$ is nonzero, then there is some $w \in W^\perp$ which is nonzero and has $\langle e_i, w \rangle = 0$ for all $e_i \in E'$, which since $E \subseteq W$ (make sure you understand why) shows that the same is true for the $v_i$, contradicting our condition on $E$. So $E'$ does indeed form a basis.

$\endgroup$
0
$\begingroup$

It's not clear what you are asking. You say we have "some set {v_1, v_2, …, v_n}" but you do not say if these vectors are independent or span the space. If they are either independent or span the space then they are a basis. You say "for v in V". Do you intend "for every v in V" or "for some v in V". Taking V to be $R^3$, we can take that set to be {(1, 0, 0), (2, 0, 0), (3, 0, 0). Then any vector v= (0, a, b) satisfies that condition. It can be true for every v only if $v_i= 0$ for all i.

$\endgroup$
  • $\begingroup$ Yes, I meant for every v in V. Nothing is said about E (neither if it is LI nor if it spans V). $\endgroup$ – daniel.franzini Aug 24 '18 at 20:05
  • $\begingroup$ If $<v,v_i>= 0$ for every v then every $v_i$ must be 0! $\endgroup$ – user247327 Aug 26 '18 at 0:28
0
$\begingroup$

Suppose $\{v_1,\dots,v_n\}$ not linearly independent. Then we can find a nonzero vector $v$ in the orthogonal complement of $\operatorname{span}\{v_1,\dots,v_n\}$. By definition of orthogonal complement, $\langle v,v_i\rangle=0$ for $i=1,2,\dots,n$.

Conversely, suppose $\{v_1,v_2,\dots,v_n\}$ is a basis for $V$. Suppose also that $\langle v,v_i\rangle=0$, for $i=1,2,\dots,n$. Then $v=a_1v_1+\dots+a_nv_n$ and $$ \langle v,v\rangle=\sum_{i=1}^n a_i\langle v,v_i\rangle=0 $$ Therefore $v$ is orthogonal to itself, so $v=0$.

As you see, there's no need that the basis is orthogonal.

$\endgroup$

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.