Showing a function forms a vector space.

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The equation 4x - y = 7 does not constitute a vector space because it is not closed under addition. When adding two points from the equation, the result yields 4(x_1 + x_2) - (y_1 + y_2) = 14, which does not satisfy the original equation. Similarly, scalar multiplication does not hold, as shown by the example with the scalar 3, leading to 3(4x_1 - y_1) = 12x_1 - 3y_1, which does not equal 3 times the original constant. The discussion clarifies that the correct interpretation involves the set S = {(x,y) | 4x - y = 7}, which is not a vector space due to these properties. Understanding these definitions is crucial for further studies in linear algebra.
schlynn
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Homework Statement


Does the function: 4x-y=7 constitute a vector space?


Homework Equations


All axioms relating to vector spaces.


The Attempt at a Solution



x_n for example means x with the subscript n

The book says that the function isn't closed under addition. So it continues by showing that given 2 points, (x_1,y_1) and (x_2,y_2) that when you add 4x_1-y_1=7 and 4x_2-y_2=7 you get 4(x_1+x_2)-(y_1+y_2)=14, how did they get the values for the problem to see that it sums to 14 and not 7? The case for multiplication show: 4x_1-y_1=7, they used 3 as the scalar to show: 3(4x_1-y_1)=12x_1-3y_1, that part makes sense, but then again they say that the right side is 3x7, and I don't see how they got those values.
 
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schlynn said:

Homework Statement


Does the function: 4x-y=7 constitute a vector space?


Homework Equations


All axioms relating to vector spaces.


The Attempt at a Solution



x_n for example means x with the subscript n

The book says that the function isn't closed under addition. So it continues by showing that given 2 points, (x_1,y_1) and (x_2,y_2) that when you add 4x_1-y_1=7 and 4x_2-y_2=7 you get 4(x_1+x_2)-(y_1+y_2)=14, how did they get the values for the problem to see that it sums to 14 and not 7?
You have ##4x_1-y_1 = 7## and ##4x_2-y_2=7##. What do you get when you add those two equations?

The case for multiplication show: 4x_1-y_1=7, they used 3 as the scalar to show: 3(4x_1-y_1)=12x_1-3y_1, that part makes sense, but then again they say that the right side is 3x7, and I don't see how they got those values.
 
You get 4(x_1+x_2)-(y_1+y_2)=14, which obviously isn't equal to 7, but does that then mean that all linear function ax+by=c is not a vector space for all c not equal to 0? Because c+c is always not equal to zero except when c=0 right?
 
Yup.
 
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Oh, ok, thanks, was just over-thinking it I guess then, it's apparent now, thank you.
 
schlynn said:

Homework Statement


Does the function: 4x-y=7 constitute a vector space?

You certainly weren't overthinking when you phrased your question. ##4x-y=7## is an equation, not a function. And a function doesn't constitute a vector space. As you proceed in your course you are going to have to be more careful with definitions so you know what you are trying to prove or disprove. Perhaps you meant to ask something like: Does the set ##S=\{(x,y)|4x-y=7\}## consitute a vector space with the usual operations?
 
LCKurtz said:
As you proceed in your course you are going to have to be more careful with definitions so you know what you are trying to prove or disprove. Perhaps you meant to ask something like: Does the set ##S=\{(x,y)|4x-y=7\}## consitute a vector space with the usual operations?
I had the same thoughts.
 
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