Dimension of a subspace and its basis. Let $V$ be an $n$-dimensional vector space and let $(a_1,a_2,\ldots,a_n)$ be a fixed vector in $V$. Prove that the collection of elements $A=(x_1,\ldots,x_n)$ of $V$ with $a_1x_1+\cdots+a_nx_n=0$ is a subspace of $V$. Determine the dimension of the subspace and find its basis.
I've solved the part related to proving that $(x_1,\ldots,x_n)$ is a subspace of $V$.
$A$ is a subspace of $V$ then we can write $\dim(A)\le\dim(V)$
The dimension of $A$ is also equal to the cardinality of its basis.
But i have no clue what to do from here.
If anyone would point me in the right direction or give me a hint, I would be grateful!
 A: If
$$(a_1,a_2,\cdots,a_n)=(0,0,\cdots,0)$$
then $\forall v \in V$, the condition holds and hence the dimension of the subspace is $n$ and you can just choose the standard basis.
If $$(a_1,a_2,\cdots,a_n) \ne (0,0,\cdots,0)$$, then $\exists a_i\ne 0$.
Therefore you can write
$$x_i=-\frac{a_1x_1+a_2x_2+\cdots+a_{i-1}x_{i-1}+a_{i+1}x_{i+1}+\cdots+a_nx_n}{a_i}$$
Therefore for all $v$ in the subspace, we have
$$v=(x_1,x_2,\cdots,x_{i-1},-\frac{a_1x_1+a_2x_2+\cdots+a_{i-1}x_{i-1}+a_{i+1}x_{i+1}+\cdots+a_nx_n}{a_i},x_{i+1},\cdots,x_n)$$
$$=\sum_{j\ne i}x_jv_j$$
where
$$v_j=(0,0,\cdots,1,\cdots,0,-\frac{a_j}{a_i},0,\cdots,0)$$
It is obvious that the $v_j$'s are linearly independent, and the number is $n-1$.
So the dimension of the subspace is $n-1$, and the $v_j$'s is a possible basis set.
A: Your space seems to be $V=\mathbb{R}^n$. Consider the linear map
$$
T\colon\mathbb{R}^n\to\mathbb{R}
\qquad
T(x_1,\dots,x_n)=a_1x_1+\dots+a_nx_n
$$
Then, by definition, $A=\ker T$.
If $(a_1,\dots,a_n)=(0,\dots,0)$, then $T$ is the zero map and its kernel is $\mathbb{R}^n$.
Otherwise, $T$ is surjective and you can apply the rank-nullity theorem

  $\dim A=n-1$

A: You don't appear to have given the vector space a name, so I'm going to call it $\ker a$.
Now either $a=0$, in which case everything is in $\ker a$, so we can take the usual basis to be a basis for $\ker a$, or $a$ has some nonzero coordinate, which I will assume is $a_n$.
Then consider the vectors $f_i = (0,\ldots,0,a_n,0,\ldots,0,-a_i)$, where $a_n$ is in the $i$th coordinate. By construction $f_i\cdot a = a_na_i - a_i a_n =0$, so $f_i\in \ker a$, and because $f_i$ is the only vector with a nonzero $i$th coordinate, all of the $f_i$ are linearly independent.  However the dimension of $\ker a$ is $n-1$, since the rank of $a$ as a linear map from $V$ to $k$ where $k$ is our base field is 1, so the nullity of $a$, which is the $\dim \ker a$ must be $n-1$ by rank-nullity. Thus our $n-1$ linearly independent vectors must in fact be a basis. Hence the $f_i$ give a basis for $\ker a$.
A: The dimension is given by the number of free variables. 
Let's suppose $\vec a\neq \vec 0$, in this case you only have one constrain:
$$a_1x_1+\cdots+a_nx_n=0$$
thus the dimension is $n-1$.
To find a basis you can set:
$x_1=a_2$ and $x_2=-a_1$ and $x_i=0$ $i=3,n$
and so on, in this way you can construct a set of $n-1$ linearly indipendent vectors which satisfy the constrain, that is a basis.
As an example you may consider a plane in $\mathbb{R}^3$ defined by: $$ax+by+cz=0$$
