Let $a\in \mathbb{R}^n$ be a nonzero vector and let $S=\{\mathbf{u:a^{t}u=0}\}$ Show that S is a subspace. What is the dimension of S? Find a matrix A whose columns form a basis of S. In particular, do this exercise for $\mathbf a=(1,-1,1)^{t}$.
Can anyone give me hint to complete my solution, what is the matrix A and what is the purpose of considering a single point a in $\mathbb{R}^3 $ in the end of the question?
My try:
Since $S=\{\mathbf{u:a^{t}u=0}\}$
$\Rightarrow \mathbf{a^{t}u=0}$
$\Rightarrow \mathbf{a_1u_1 +a_2u+a_3u_3+...+a_nu_n=0}$
Since a is the non zero vector so I think the only solution for above equation is that all $\mathbf{u_i =0 }$.
Hence the set $S=\{\mathbf{0}\}$. And it is subspace(we can prove it).
So the basis of the $S \ is \ \{\mathbf{}\}$ and the dimension of S is the 0.
What will be the possible answers of the remaining questions?