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

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 $n$.

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Not orthogonal? Take $v_1$. – Git Gud Feb 10 at 22:09
Why is $v_1$ not orthogonal to any of the other vectors? – user61799 Feb 10 at 22:11
Since the vectors are linearly independent, $\|v_1\|>0$. Then $v_1\cdot v_1 = \|v_1\|^2 > 0$, hence $v_1$ is not orthogonal to itself. – user7530 Feb 10 at 22:12
@GitGud $(1,0),(0,1)$. However, the statement $(1,0)$ is not orthogonal to both vectors is false. – Amr Feb 10 at 22:14
"orthogonal" or "not orthogonal"? The title is wrong? – leonbloy Feb 10 at 22:36
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 $V$.
 Actually, there is no need to assume $k 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$Vwe require \begin{align} \langle v_j|u\rangle&=\sum_{i=1}^k \langle v_j| v_i\rangle a_i\\ &\neq 0. \end{align} The above is a set ofk$inequalities with$k$unknowns, which can be solved in principle. A vector of the desired form exists. - 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. - 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)\$.