orthogonal subspace - basis

Question

Let U be sub-space of $\mathbb{R}^n$ that is defined:

$U=\{(x_1,...,x_n)\in \mathbb{R}^n\mid\sum_{i=1}^{n} x_i=0\}$

So ${(1,1,...,1)}$ is a basis of $U^⊥$.

I need to prove this.

Any hints?

Answer 1

Hint:

Observe that $\;\dim U=n-1\;$ ( you can prove this by showing that $\;U=\ker\phi\;$ , with $\;\phi\in(\Bbb R^n)^*=\;$ the dual of $\;\Bbb R^n\;$) , and thus any single vector in $\;U^\perp\;$ is a basis of it...

Answer 2

Here is a direct proof without invoking any theorem of linear algebra: If $(x_1,x_2,...,x_n) \in U$ then $(x_1,x_2,...,x_n)$ is orthogonal to $(1,1,...,1)$ by definition. Suppose $(a_1,a_2,...,a_n)$ is a vector orthogonal to $U$. For any two indices $i$ and $j$ consider the vector $x$ with 1 at the i-th place and -1 at the j-th place and 0 elsewhere. Then $x \in U$ and orthogonality of $(a_1,a_2,...,a_n)$ with this vector gives $a_i=a_j$. Since this is true for all $i$ and $j$ we see that $(a_1,a_2,...,a_n)$ is a multiple of $(1,1,...,1)$. Hence $(1,1,...,1)$ spans the orthogonal complement of $U$.