Find a linear functional such that vanishes on given vectors

Question

Suppose that $v_1,v_2,...,v_k\in\mathbb{R}^n\setminus \{0\}.$ Show that there exists a linear functional $f:\mathbb{R}^n\rightarrow \mathbb{R}$ such that $f(v_j)\neq 0$ for all $j.$

I just taking $f(x)=\langle x,y\rangle$ for fixed $y\neq 0 $. This map is linear. But I am not sure whether it will work.

Your choice will not work. Consider a vector orthogonal to y. Unless you construct a y that is not orthogonal to any of your v1,...,vk. — Mark, Nov 18 '14 at 05:31
Suppose you have just one vector $v$.. Can you find a linear functional $f$ such that $f(v)\neq 0$? — , Nov 18 '14 at 05:34
first think about single vector.. then try for more than one vector.. What linear functionals are you aware in case of $\mathbb{R}^n$ to $\mathbb{R}$.. — , Nov 18 '14 at 05:39
Hey.. Common... How is this different from $f(x)=\langle x,y\rangle$.. — , Nov 18 '14 at 05:51
This isn't different, but you have to actually pick a good $y$. Not all choices will work. — Kevin Carlson, Nov 18 '14 at 05:59

score 1 · Answer 1 · edited Apr 13 '17 at 12:21

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The set of vectors orthogonal to $v_1$ forms a subspace $V_1$. Because $v_1\neq0$ the subspace $V_1$ is not all of $\Bbb{R}^n$. Similarly for all $i$, the set of vectors orthogonal to $v_i$ forms a proper subspace $V_i$, $i=1,2,\ldots,k$.

At this point I refer you to the latter part of this answer, where I show that the union $\bigcup_{i=1}^kV_i$ is not all of $\Bbb{R}^n$ either. Can you do the rest with the inner product idea?

The argument works even if we had countably infinitely many vectors $v_i,i\in\Bbb{N}$.

edited Apr 13 '17 at 12:21

Community

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answered Nov 18 '14 at 06:11

Jyrki Lahtonen

140,891

The same argument later appeared here. This link may be better for you because the parts dealing with finite fields was left out. Leaving the link to the older version into the answer. – Jyrki Lahtonen Nov 18 '14 at 06:15

Find a linear functional such that vanishes on given vectors

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