Show that the dimension of the solutions of the equation $a_1x_1+a_2x_2+...+a_nx_n=0$ is $n-1$. Show that all subspaces of the dimension $n-1$ has this same form.
I don't know even where to begin. Here are just some thoughts: The $a_1x_1+a_2x_2+...+a_nx_n=0$ can be represented as a product of two matrices: the matrix of $a$'s $n\times 1$ and the one of $x$'s $1\times n$. So, I guess the dimension of the $a_1x_1+a_2x_2+...+a_nx_n=0$ must be the dimension of the $n\times 1$ matrix of $a$'s. Its dimension can be at most $n-1$.
You are on the right track: you have $A\mathbf x=0$ for the matrix $A=[a_1,\cdots ,a_n]$. And you are assuming that at least one of the $a_i$ s non-zero. Now, $P=\{\mathbf{x}\in \mathbb R^n \mid A\mathbf{x}=0\};$ that is, $P$ is the nullspace of $A$. And since at least one of the $a_i$ is non-zero, the rank of $A$ is $1$. Now the result follows from the rank-nullity theorem, because $\text{rk}A +\text{ker}A =n$.
A faster way would be to note that $P$ is just the orthogonal complement of the one dimensional vector space generated by $(a_1,\cdots, a_n)$ and so has dimension $n-1.$
I leave the converse to you.