Suppose $a$ is a scalar and $\vec{v}$ is a vector, $a\vec{v}=0$ implies that $a=0$ or $\vec{v}=0$

$$\text{Suppose }a \text{ is a scalar and } \vec{v} \text{ is a vector in } V, a\vec{v}=0 \text{ implies that } a=0 \text{ or } \vec{v}=0$$

I knew the $$a\ne 0$$ then $$v=0$$ proof. I asked other things about the statement and then he tried to prove it by contraposition. I was a bit confused about his proof, but didn’t have enough time to discuss with him.

Below is his proof.

Proof:
prove by contraposition. The contrapositive statement is “Suppose $$a$$ is a scalar and $$\vec{v}$$ is a vector in $$V$$, $$a\ne 0$$ and $$\vec{v}\ne 0$$ imply that $$a\vec{v}\ne 0$$.

Let $$v\in V$$ and $$\{e_1, e_2, ..., e_n\}$$ be a basis of $$V$$, then we can write $$v= \begin{bmatrix} v_1\\ v_2\\ ...\\ v_n \end{bmatrix}$$ regarding the basis. (I ask what if $$V$$ is infinite-dimensional, and he said some infinite vector spaces have a basis but it’s unproven whether all have a basis, so we assume there is a basis for all (I forget what he exactly said, as I was a bit confused))

$$a\vec{v} = a \begin{bmatrix} v_1\\ v_2\\ ...\\ v_n \end{bmatrix}= \begin{bmatrix} av_1\\ av_2\\ ...\\ av_n \end{bmatrix}$$

If $$\vec{v}\ne 0$$, there exists $$v_i\ne 0$$ such that $$av_i\ne 0$$ (I think this is true if $$v_i$$ is a scalar, but $$v_i$$ doesn’t have to be a scalar as $$V$$ is a random vector space, right?), then at least an entry of $$a\vec{v}$$ isn’t zero, $$a\vec{v}\ne 0$$.

Then finish.

Words in bold are my questions, and I wonder whether this proof also applies to infinite-dimensional $$V$$ as he said. Is there a better contrapositive proof for this one? I know the contradiction proof which is way easier, but wonder about a contrapositive proof as he already mentioned it.

• Suppose $\alpha v = 0$ and $\alpha \neq 0$. Then, we can multiply both sides by $\alpha ^{-1}$ and deduce that $v = 0$. Thus $\alpha = 0$ or $v = 0$. Commented Sep 11, 2020 at 23:58
• By accepting the axiom of choice, every infinite dimensional vector space has a basis. However, it may be so ugly that it is practically useless. Regardless, the proof you outlined still works. Commented Sep 12, 2020 at 0:16

This simple proof applies even if the vector space has infinite dimension. Let us recall that all $$K$$-vector spaces satisfy the Identity element of scalar multiplication, that is $$1.\vec{v}=\vec{v}$$, where $$1\in K$$.
Then if $$a=0$$ the assertion if obvious, so let me suppose $$a\neq 0$$, then you can multiply $$a.\vec{v}=0$$ to both sides by $$\frac{1}{a}$$, then you have $$\frac{1}{a}.(a.\vec{v})=0$$ and if we apply here the compatibility of scalar multiplication with field multiplication obtain $$(\frac{1}{a}a).\vec{v}=0$$ or $$1.\vec{v}=0$$, but we know that the left hand side is just $$v$$, then $$\vec{v}=0$$.
A basis isn't required to show this. First, consider a scalar $$\alpha \neq 0$$ then if $$\alpha v = 0$$ we multiply on the left by $$\alpha^{-1}$$ to arrive at $$v=0$$.
Now if $$\alpha = 0$$ then we note that $$v+v = 2v$$ and so $$0v + 0v = 2(0v)=0v$$ and now add $$-0v$$ to both side of $$0v + 0v = 0v$$ gives us $$0v=0$$ as desired.