This question from linear algebraSuppose you [ have a consistent system of linear equations, with...

It is given that we have a consistent system of linear equations with coefficients in ${\displaystyle \mathbb{R}}$. So, the system is written as Ax=0 which is a homogeneous system with all the bi equal to 0 when Ax=b.
We know that the system Ax=0 is said to be consistent if it has a solution. We have to show that the set of solutions of this homogeneous system form a vector space over ${\displaystyle \mathbb{R}}$.
Property 1: The set of solutions contains the zero vector.
If we take the vector x as a zero vector i.e., x=0 then A(0)=0. This implies that the set of solutions contain the zero vector.
Property 2: If ${\displaystyle x,y\in V}$ then ${\displaystyle x+y\in V}$ where V is the set of solutions of the given system.
Let x and y are two solutions of the homogeneous system. Then, we have, Ax=0 and Ay=0.
Now,
${\displaystyle A(x+y)=Ax+Ay}$=0+0=0$\therefore A(x+y)=0$
Hence, x+y is a solution of the homogeneous system which implies that the set of solutions is closed under addition.
Property 3: If c is any scalar and x is a solution of the homogeneous system, then cx is also a solution of the homogeneous system. Since x is a solution of the homogeneous system, Ax=0.
Now,
${\displaystyle A\left(cx\right)=c\left(Ax\right)=c\left(0\right)=0\therefore A\left(cx\right)=0}$
Hence, cx is a solution of the homogeneous system.
Therefore, the set of solutions of the homogeneous system form a vector space over ${\displaystyle \mathbb{R}}$.
The set of solutions will not form a vector space if we take the non-homogeneous system i.e., Ax=b where b contains atleast one nonzero element.
Reason: If we take x=0,
${\displaystyle A\left(0\right)=b}$
${\displaystyle b=0}$
but b is nonzero which implies that ${\displaystyle A\left(0\right)\ne 0}$. Therefore, the set of solutions of the non-homogeneous system does not contain the zero vector and hence, it is not a vector space.