Given $n$-linearly independent vectors how do I find a vector not orthogonal to any of them?

Question

I'm trying to do this as part of another proof:

Let $v_1, \ldots, v_k \in \mathbb{R}^{n}$ be linearly independent vectors. How do I find a vector that's not orthogonal to any of these?

Edit: the proof doesn't necessarily have to be constructive. I just need to know that such a vector exists.

Edit 2: I just realized that linear independence doesn't need to hold. Then $k$ is allowed to be greater than $n$.

Answer 1

Let $V=[v_1,\ldots,v_k]$ (so that $V$ is $n\times k$) and $\mathbf{1}$ be the $k$-vector with all entries equal to $1$. Since the rank of $V$ is $k$, the equation $V^Tx=\mathbf{1}$ has a solution, and this $x$ is not orthogonal to any column of $V$.

Answer 2

I will be using Dirac notation to address the question.

Consider a set of linearly independent vectors $$V=\{|v_i\rangle : i \in \{1,\ldots,k\}\}$$ and a linear combination thereof given by $$|u\rangle=\sum_{i=1}^k a_i |v_i\rangle,$$ where $a_i$ are expansion coefficients.

For $|u\rangle$ to have a non-zero projection onto any of the vectors in $V$ we require $$\begin{align} \langle v_j|u\rangle&=\sum_{i=1}^k \langle v_j| v_i\rangle a_i\\ &\neq 0. \end{align}$$

The above is a set of $k$ inequalities with $k$ unknowns, which can be solved in principle. A vector of the desired form exists.

Answer 3

Let $V_1, \dots, V_k$ be the one-dimensional subspaces spanned by $v_1, \dots, v_k$ respectively. Then what you're asking is equivalent to: prove that $\mathbb{R}^n$ is not equal to $V_1 ^{\perp} \cup \dots \cup V_k ^{\perp}$, where $V_i ^{\perp}$ denotes the orthogonal complement subspace to $V_i$. This follows from the fact that this union can't be a subspace since the vectors are distinct.

Answer 4

Let $A$ be the $n\times k$ matrix whose columns are the vectors $v_1,\ldots,v_k$. By assumption, it has rank $k$. So $A^t$ has rank $k$ and is $k\times n$. Hence it admits a left $k\times k$ inverse, say $B$.

Now pick your favorite vector $b$ with nonzero coefficients, say $b=(1,\ldots,1)$, in $\mathbb{R}^k$.

We want to solve for $v\in\mathbb{R}^n$ such that $$ A^tv=b. $$ This yields $$ v=Bb. $$

The vector $v$ answers your problem, since the coordinates of $A^tv$ are precisely the inner products $(v_j,v)$.