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How can I find a basis for the subspace $V:=\{v = (v_1, v_2, \cdots , v_n) \in R^n: v_1+v_2+ \cdots +v_n=0\}$ of $R^n$ for any $n$? I know that I must show that the basis is linearly independent and spans $V.$

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2 Answers 2

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It is the nullspace of the $1\times n$ matrix

$$\left[ 1\ \ 1 \ \ 1 \ \ \cdots \ \ 1\right]$$

Which has $n-1$ free variables. Can you finish?

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First, notice that your defining equation actually describes a hyperplane i $\mathbb{R}^n$ of dimension $n-1$ - in fact, you will prove this when proving that the space has a basis of dimension $n-1$.

An example of such a basis of this space is the following: $(1,-1,0,\ldots,0), (0,1,-1,0,\ldots,0), (0,0,1,-1,0,\ldots,0), (0,0,\ldots,0,1,-1)$.

The remaining job is then to show that this set of vectors is actually linearly independent and spans $V$.

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  • $\begingroup$ Would I use Gaussian elimination to prove linear independence and spans V? How can I show this for all n? $\endgroup$
    – user187039
    Commented Jan 13, 2015 at 19:24
  • $\begingroup$ I'd say, take an arbitrary vector $v=(v_1,\ldots,v_n)$ in your space $V$. To show that the basis I gave above spans $V$, you should be able to write $v$ as a linear combination of the basis vectors. Can you see how to do this? Also, to show that my given set of vectors is linearly independent, try taking a linear combination of them, and equate this linear combination to 0. Can you infer that all coefficients must be 0? $\endgroup$
    – Mankind
    Commented Jan 14, 2015 at 19:54
  • $\begingroup$ Thanks for all of your help- I have managed it now! $\endgroup$
    – user187039
    Commented Jan 16, 2015 at 11:45

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