TITLE: A Scalar times the Zero Vector
QUESTION [5 upvotes]: I'm reading Linear Algebra Done Right by Sheldon Axler and the proof given in the book is the same as the one in the answer provided for this question.
I tried to solve this before looking at the solution and the way I did it was:

Theorem: $a \cdot \vec0 = \vec 0 $ for every $a \in \mathbb F$ 

Proof $\ $Let $a \in \mathbb F$, then 
\begin{align}a \cdot \vec0    
&= a \cdot \langle 0_1,0_2, \ldots ,0_n\rangle \tag{Def. of a vector}\\
&= \langle a \cdot0_1,a \cdot0_2, \ldots ,a \cdot0_n \rangle \tag{Def. of Scalar Multiplication} \\
&= \langle 0_1,0_2,...,0_n \rangle \\
&= \vec 0
\end{align}
Hence,  $a \cdot \vec0 = \vec 0 $ , desired result.

Is there anything wrong with this proof? For example, I didn't explain why $a \cdot 0_j = 0$, do I have to do so? 
Also doesn't this proof provide more insight in terms of using basic definitions rather than just vector algebra?*
Is there a way to proof this result besides this and the one given in the link?

REPLY [4 votes]: Use the fact that the zero vector is the neutral element for vector addition, so it is equal to itself minus itself. Then use the distributive law for scalar multiplication.