How to prove that the solutions of a linear system Ax=0 is a vector space over R?

Question

How to prove that the set of solutions to a homogeneous linear system $Ax=0$ is a vector space over $R$ under usual addition and scalar multiplication? Can you guys provide me an answer for this?

Answer 1

Hint: You seem to be getting hung up on the definitions. Keep in mind that you should be answering the following:

• For vectors $x$ and $y$: if $Ax = 0$ and $Ay = 0$, why should we believe that $A(x + y) = 0$?
• For a vector $x$ and scalar $\alpha$: if $Ax = 0$, why should we believe that $A(\alpha x) = 0$?

Once you have an answer for the above two questions, you will have proven that the set of all vectors satisfying $Ax = 0$ is "a vector space over $\Bbb R$ under the usual addition and scalar multiplication".

Answer 2

To show something that $W$ is a subspace of $V$, we must show that:

1) $W$ is non-empty

2) $W$ is closed under vector addition and scalar multiplication. This is:

$\forall x,y \in W: \forall a \in \mathbb{R}: x + y \in W, ax \in W$

So, let's check these things:

1) $A*0 = 0$, so $0 \in \{Z|AZ = 0\}:= K$. Hence $K \neq \emptyset$

2) If $X,Y \in K$, then $A(X+Y) = AX + AY = 0 + 0 = 0$ hence $X + Y \in K$

Also $A(aX) = a(AX)= a*0 = 0$, so $aX \in K$

Hence, $K$ is a subspace.