If $x \not= 0$ and $\lambda x = 0$, then $\lambda = 0$.

Question

I am trying to use the definition of vector spaces to prove that, if $x \not= 0$ and $\lambda x = 0$, then $\lambda = 0$.

Expressing the proposition differently, it seems to me that it is saying the following: If $x \not= 0$ and $\lambda x = 0$, then there exists some $\lambda \in E$ such that $\lambda x = 0$.

Let $x \not= 0$ and $\lambda x = 0$, where $x, \lambda \in E$.

Therefore, by the definition of vector spaces, there exists some $w \in E$ such that $x + \lambda = w$.

Since $x, w \in E$, let $w = x$ (that is, we select $w$ to be equal to $x$).

$\therefore x + \lambda = w \Rightarrow \lambda = w - x = 0 \ \ \ Q.E.D.$

I would greatly appreciate it if people would please take the time to clarify whether or not this is correct.

Answer 1

The sentence “If $x\ne0$ and $\lambda x=0$, then there exists some $\lambda\in E$ such that $\lambda x=0$.” makes no sense; what is the first $\lambda$?

And the problem of proving that $x\ne0$ and $\lambda x=0$ implies that $\lambda=0$ contains the implicit assumption that $x$ is a vector and that $\lambda$ is a scalar. But then $x+\lambda$ is meaningless.

You can prove what you want to prove as follows: if $\lambda\ne0$, then\begin{align}0&=\lambda^{-1}0\\&=\lambda^{-1}(\lambda x)\\&=\left(\lambda^{-1}\lambda\right)x\\&=1x\\&=x.\end{align}But we are assuming that $x\ne0$.

Answer 2

Lemma: For the vector $0$ and any scalar $\lambda$ we have $\lambda 0 = 0$:

Define $z=\lambda 0$, then

• $z=\lambda 0 = \lambda(0+0)$ (as $0= 0+0$ in the vector space)
• distributivity tells us $\lambda(0+0)=\lambda 0 + \lambda 0 =z +z$.
• combining we get $z = z+z$.
• Then add $-z$ to both sides to get $0 = (z + (-z)) = (z+ z) + (-z)$.
• now, associativity for the right hand side gives us $0 = z + (z + (-z)) = z + 0 = z$.
• So $z=0 = \lambda 0$ and we're done.

Knowing that, suppose $\lambda x = 0$ for some $\lambda$ and $x \neq 0$. Now assume $\lambda \neq 0$ so that $\lambda^{-1}$ exists (as a scalar) such that $\lambda^{-1} \lambda=1$ (here we use that scalars are a field).

Then:

• $x = 1 x$ (axiom) so that:

• $x = (\lambda^{-1} \lambda)x$, using another axiom we get

• $x= \lambda^{-1}(\lambda x) = \lambda^{-1} 0 = 0$, where we substitute $\lambda x = 0$ and use the previous lemma for the scalar $\lambda^{-1}$.

So $x=0$, contradiction with $x \neq 0$, so the assumption $\lambda \neq 0$ was wrong and $\lambda = 0$. QED.