Proving Vector Space Property: αa = 0 ⟹ α = 0 or a = 0

In summary, we can prove that in any vector space V, ##\alpha \overrightarrow a = \overrightarrow 0## implies that either ##\alpha = 0## or ##\overrightarrow a = \overrightarrow 0## by distinguishing cases and using the distributive property of logical operators.
  • #1
member 587159

Homework Statement



Prove that in any vector space V, we have:

##\alpha \overrightarrow a = \overrightarrow 0 \Rightarrow \alpha = 0 \lor \overrightarrow a = \overrightarrow 0##

Homework Equations



I already proved:

##\alpha \overrightarrow 0 = \overrightarrow 0##
##0 \overrightarrow a = \overrightarrow 0##

The Attempt at a Solution


[/B]
Suppose ##\alpha \neq 0##

Then: ##\alpha \overrightarrow a = \overrightarrow 0##
##\Rightarrow \alpha^{-1} (\alpha \overrightarrow a) = \alpha^{-1} \overrightarrow 0##
##\Rightarrow (\alpha^{-1} \alpha) \overrightarrow a = \overrightarrow 0##
##\Rightarrow 1 \overrightarrow a = \overrightarrow 0##
##\Rightarrow \overrightarrow a = \overrightarrow 0##

The problem is. I don't know how to show that ##\alpha \overrightarrow a = \overrightarrow 0## can imply ##\alpha = 0## I can't suppose ##\alpha = 0##, because I need to prove that?

Maybe something like this?

##\alpha \overrightarrow a = \overrightarrow 0##
But ##0 \overrightarrow a = \overrightarrow 0##

Thus: ##\alpha \overrightarrow a = 0 \overrightarrow a ##

Comparing the two, we obtain ##\alpha = 0##

Thanks in advance.
 
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  • #2
I suggest you focus on the implications of assuming that ##\vec a \neq 0##.
 
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  • #3
You can distinguish cases. Either ##\alpha = 0## or ##\alpha \neq 0##. One of the two has to happen.
 
  • #4
fresh_42 said:
You can distinguish cases. Either ##\alpha = 0## or ##\alpha \neq 0##. One of the two has to happen.

But can I assume ##\alpha = 0##? Then it follows trivially that ##\alpha = 0 \land \alpha \overrightarrow a = \overrightarrow 0 \Rightarrow \alpha = 0##
 
  • #5
Math_QED said:
But can I assume ##\alpha = 0##? Then it follows trivially that ##\alpha = 0 \land \alpha \overrightarrow a = \overrightarrow 0 \Rightarrow \alpha = 0##
Why not?
$$A = A \wedge \text{ true } = A \wedge (B \vee \lnot B) = (A \wedge B) \vee (A \wedge \lnot B)$$
and ##B=(\alpha = 0)## does the job.
 
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  • #6
fresh_42 said:
Why not?
$$A = A \wedge \text{ true } = A \wedge (B \vee \lnot B) = (A \wedge B) \vee (A \wedge \lnot B)$$
and ##B=(\alpha = 0)## does the job.

Nice. Thanks a lot.
 

1. What does it mean for a vector space property to be proved?

Proving a vector space property means demonstrating that the given property holds true for all elements in a vector space. This involves showing that the property is satisfied by any arbitrary vector in the space, using a rigorous mathematical approach.

2. What is the significance of the property αa = 0 ⟹ α = 0 or a = 0 in vector spaces?

This property, also known as the "zero product property", is essential in determining the linear independence of vectors in a vector space. It states that if a vector multiplied by a scalar equals zero, then either the scalar or the vector must be zero.

3. How is this property proven in vector spaces?

To prove this property, we must show that it holds true for any arbitrary vector a and scalar α in a vector space. This can be done using the axioms and definitions of vector spaces, such as closure under scalar multiplication and the existence of a zero vector.

4. What implications does this property have on vector operations?

The zero product property is important in understanding vector operations, as it allows us to determine when a vector can be expressed as a linear combination of other vectors. It also helps to identify when vectors are linearly dependent or independent.

5. Are there any exceptions to this property in vector spaces?

No, this property holds true for all vector spaces. It is a fundamental property that is necessary for a set to be considered a vector space. Any set that does not satisfy this property cannot be classified as a vector space.

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