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

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In a vector space V, the statement "if cx=0 then c=0" is only valid when x is non-zero. To prove this, one can assume c is not zero and multiply both sides of the equation cx=0 by the inverse of c, leading to a contradiction that x must equal zero. The proof relies on the cancellation law and the uniqueness of the zero vector. The discussion emphasizes the importance of specifying that x cannot be zero for the proposition to hold true. Therefore, the assertion is confirmed under the condition that x is non-zero.
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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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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.
 
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.
 
Good!

Actually you get: c-1cx=c-10, therefore 1x=x=0. This is use of the cancellation law.
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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