# If $\langle v,v_i \rangle = 0$ then $v=0$, what can I say about the span of $v_i$?

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.

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.

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.

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

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.