Proving Subspace in R^3 with Fixed Matrix A: W={x \in R^{3} :Ax=[^{1}_{2}]}

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SUMMARY

The discussion focuses on proving that the set W = {x ∈ R³ : Ax = [1; 2]} is a subspace of R³, where A is a fixed 2x3 matrix. Participants clarify that the notation after the colon indicates a condition that must be satisfied for vectors x to belong to the set W. The conversation emphasizes the importance of understanding set notation and theorems related to subspaces in linear algebra.

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  • Understanding of linear algebra concepts, particularly subspaces.
  • Familiarity with matrix operations and notation.
  • Knowledge of set notation and its implications in mathematical contexts.
  • Proficiency in the properties of vector spaces in R³.
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  • Study the criteria for a set to be classified as a subspace in linear algebra.
  • Learn about the properties of linear transformations and their impact on vector spaces.
  • Explore examples of proving subspaces using specific matrices and conditions.
  • Review theorems related to the intersection of subspaces in R³.
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Students and educators in mathematics, particularly those focusing on linear algebra, as well as anyone involved in theoretical proofs concerning vector spaces and subspaces.

eyehategod
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Let A be a fixed 2x3 matrix. Prove that the set
[itex]W={x \in R^{3} :Ax=[^{1}_{2}]}[/itex] (2x1 matrix 1 on top 2 at the bottom)


what does the information after the ":" mean? is it a condition?
I don't understand this problem. Can anyone help me out?
 
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eyehategod said:
what does the information after the ":" mean? is it a condition?
I don't understand this problem. Can anyone help me out?

Yes, it is a condition. You're trying to prove if this set of vectors x such that the condition after ":" holds is a subspace of R^3, I guess? You didn't completely lay out the problem, btw.
 
Ok, this is how set notation works.

If I say...

A = { x : x e N }

Then that read... x is an element of A if x is an element N (the natural numbers). Hence, A = N, right? Do you agree?

Let's think of something a little harder...

A = { (x,y) : x + y = 1, and x,y e Z }

Z is all the integers including 0. So, what's in A? Well, all (x,y) are elements of A if x+y=1 and x,y element of Z. An example is (0,1) because 0+1 = 1 and 0 e Z and 1 e Z.

Anyways, now go back and re-read that question and state it correctly. And look for the Theorem on how to show a set is a subspace.
 

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