For x, y in a vector space V, c in F, if cx=0 then c=0?

In summary, we have discussed the proposition that for x in a vector space V, c in F, if cx=0 then c=0. This was originally from Friedman's Linear Algebra on page 12. To prove this, we can use the cancellation law and the fact that 0 and -x are unique, as well as the fact that 0x=0, in accordance with basic vector space definitions. However, this proposition is only true for non-zero x. To prove it, we can assume that c is not equal to 0 and then multiply both sides by the inverse of c, leading to a contradiction. Therefore, our original assumption must be false and c must equal 0. This can also be seen by
  • #1
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Homework Statement



For x, y in a vector space V, c in F, if cx=0 then c=0?
How do you prove this?
This is originally from Friedman, Linear Algebra, p.12.
To prove this, I can use a few facts:

(1) cancellation law
(2) 0, -x are unique
(3) 0x = 0

with basic vector space definition.

Homework Equations





The Attempt at a Solution

 
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  • #2
Welcome to PF, julypraise! :smile:

julypraise said:

Homework Statement



For x, y in a vector space V, c in F, if cx=0 then c=0?
How do you prove this?
This is originally from Friedman, Linear Algebra, p.12.
To prove this, I can use a few facts:

(1) cancellation law
(2) 0, -x are unique
(3) 0x = 0

with basic vector space definition.

Your proposition is only true for non-zero x.
Should I assume that is an additional constraint?

If so, then to prove it, I suggest the following approach.

Start with: suppose that c≠0, and then multiply with the inverse of c.
 
  • #3
Okay, thanks. Let me follow your instruction.

Prop. For x in V not equal to 0, c in F, if cx=0 then c=0.

Proof. Suppose c is not equal 0. Then if we times 1/c on the both sides of cx=0
we get 1x = c0 = 0 Therefore x = 0, which is contradiction. QED

Okay it works fairly well. Thx.
 
  • #4
Good!

Actually you get: c-1cx=c-10, therefore 1x=x=0. This is use of the cancellation law.
 

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