# Thread: Proof of a small lemma (systems of linear equations)

1. ## Proof of a small lemma (systems of linear equations)

Suppose v (a vector) is a solution to Ax=0, and Q is an element of the reals. Show Qv is also a solution.

I went about this by saying that Av = 0 (by the equation), and that A(Qv) = Av + Av + Av + Av + .... up to Q times, which = 0 + 0 + 0 + 0 + ... = 0
I don't think this is a proper way of proving this fact, though...can anyone point me in the right direction?

2. Originally Posted by mistykz
Suppose v (a vector) is a solution to Ax=0, and Q is an element of the reals. Show Qv is also a solution.

I went about this by saying that Av = 0 (by the equation), and that A(Qv) = Av + Av + Av + Av + .... up to Q times, which = 0 + 0 + 0 + 0 + ... = 0
I don't think this is a proper way of proving this fact, though...can anyone point me in the right direction?
$\displaystyle \bold v$ is a solution to $\displaystyle \bold A\bold x=\bold 0$. We need to show that $\displaystyle Q\bold v$ is also a solution, where $\displaystyle Q\in\mathbb{R}$.

Substituting $\displaystyle \bold x=Q\bold v$ into the equation, we see that $\displaystyle \bold A(Q\bold v)=0\implies Q(\bold A\bold v)=\bold 0$

Since $\displaystyle \bold A\bold v=\bold 0$, we now see that $\displaystyle Q(\bold 0)=\bold 0\implies \bold 0=\bold 0$.

We have verified that $\displaystyle Q\bold v$ is a solution to $\displaystyle \bold A\bold x=\bold 0$

$\displaystyle \mathbb{Q.E.D.}$

I hope this makes sense!

--Chris