MHB Basic equation in Vector Space - Cooperstein Exercise 1, Section 1.3

[Math Amateur]

I am reading Bruce Cooperstein's book: Advanced Linear Algebra ... ...

I am focused on Section 1.3 Vector Spaces over an Arbitrary Field ...

I need help with Exercise 1 of Section 1.3 ...

Exercise 1 reads as follows:View attachment 5107Although apparently simple, I cannot solve this one and would appreciate help ...

Peter*** EDIT ***

To give MHB readers an idea of Cooperstein's notation and approach I am providing Cooperstein's definition of a vector space ... as follows:View attachment 5108

 

[Deveno]

$0 = 0 + 0$ (A3)
$c0 = c(0 + 0)$ (multiplying both sides by $c$)
$c0 = c0 + c0$ (M1)
$c0 + (-c0) = (c0 + c0) + (-c0)$ (adding $-c0$, which exists by A4, to both sides)
$0 = (c0 + c0) + (-c0)$ (A4, again, on the LHS))
$0 = c0 + (c0 + -(c0))$ (A2, on the RHS)
$0 = c0 + 0$ (A4)
$0 = c0$ (A3).

The crux of this argument has its origins in a similar theorem from group theory:

In a group $(G,\ast)$ if $a\ast a = a$, then $a = e$. The only part of it that is "vector-y" is where we invoke M1 to write $c0$ as $c0 + c0$. This is an oft-used type of proof, for example in the real numbers, to show $x = 0$, it suffices to show that $x = 2x$, and in the non-zero reals, if $t$ is a root of $t^2 - t$, then $t = 1$ (we have to exclude $t = 0$ from the reals to get a *group* as $0$ has no multiplicative inverse).

 

[Math Amateur]

$0 = 0 + 0$ (A3)
$c0 = c(0 + 0)$ (multiplying both sides by $c$)
$c0 = c0 + c0$ (M1)
$c0 + (-c0) = (c0 + c0) + (-c0)$ (adding $-c0$, which exists by A4, to both sides)
$0 = (c0 + c0) + (-c0)$ (A4, again, on the LHS))
$0 = c0 + (c0 + -(c0))$ (A2, on the RHS)
$0 = c0 + 0$ (A4)
$0 = c0$ (A3).

Thanks Deveno ...

hmm ... easy to follow the solution ... but not so easy to dream up or create solution ... :(

Thanks again for the help ... much appreciated ...

Peter