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

Question

I asked a math graduate something about this statement:

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

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.

Thanks in advance!

Answer 1

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 $\vec{v}=0$.

I hope to be clear. 

Answer 2

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.