The discussion revolves around finding the basis for the subspace defined by the equation 4x + y - 3z = 0. Participants are exploring the concept of linear independence in the context of the proposed basis vectors.
Some participants have provided guidance on how to demonstrate linear independence, while others express uncertainty about the definitions and the implications of the linear expression versus the subspace. The discussion is ongoing, with some participants seeking further clarification on the concepts involved.
There is a noted confusion regarding the interpretation of the original problem statement, particularly the importance of the "= 0" in defining the subspace. Additionally, participants are grappling with the implications of the basis vectors and their linear independence.
That is not a subspace- it is a linear expression! I suspect that you mean "Find a basis for the subspace of R3 of all (x, y, z) satisfying 4x+ y- 3z= 0".negation said:Homework Statement
Find the basis for the subspace 4x+y-3z
The definition of "u, v are Linearly Independent" is that "if au+ bv= 0 then a= b= 0". Applied here, that would be a[1; -4; 0]+ b[0; 3; 1]= [a; -4a+ 3b; b]= [0; 0; 0] which tells you that a= 0; -4a+ 3b= 0; b= 0. The first and third equations tell you everything you need to know, don't they?The Attempt at a Solution
I found that the basis is {[1;-4;0],[0;3;1]}. How do I know if it is linearly independent? I know that the mathematical definition of what LI is but how can it be applied to show in this case?
HallsofIvy said:That is not a subspace- it is a linear expression! I suspect that you mean "Find a basis for the subspace of R3 of all (x, y, z) satisfying 4x+ y- 3z= 0".
(If you had written "Find a basis for the subspace 4x+ y- 3z= 0" I would have had no problem, interpreting it a short hand for the above. But the "= 0" is important. Also note "a basis" not "the basis". Any vector space or subspace has an infinite number of bases.)
The definition of "u, v are Linearly Independent" is that "if au+ bv= 0 then a= b= 0". Applied here, that would be a[1; -4; 0]+ b[0; 3; 1]= [a; -4a+ 3b; b]= [0; 0; 0] which tells you that a= 0; -4a+ 3b= 0; b= 0. The first and third equations tell you everything you need to know, don't they?
HallsofIvy said:The definition of "u, v are Linearly Independent" is that "if au+ bv= 0 then a= b= 0". Applied here, that would be a[1; -4; 0]+ b[0; 3; 1]= [a; -4a+ 3b; b]= [0; 0; 0] which tells you that a= 0; -4a+ 3b= 0; b= 0. The first and third equations tell you everything you need to know, don't they?
negation said:How do I know a[1;-4;0]+b[0;3;1] =[0;0;0]?
LCKurtz said:You are wanting to show those two vectors are linearly independent. What about Hall's explanation don't you understand?
negation said:Homework Statement
Find the basis for the subspace 4x+y-3z
The Attempt at a Solution
I found that the basis is {[1;-4;0],[0;3;1]}. How do I know if it is linearly independent? I know that the mathematical definition of what LI is but how can it be applied to show in this case?
HallsofIvy said:The definition of "u, v are Linearly Independent" is that "if au+ bv= 0 then a= b= 0". Applied here, that would be a[1; -4; 0]+ b[0; 3; 1]= [a; -4a+ 3b; b]= [0; 0; 0] which tells you that a= 0; -4a+ 3b= 0; b= 0. The first and third equations tell you everything you need to know, don't they?
negation said:How do I know a[1;-4;0]+b[0;3;1] =[0;0;0]?
LCKurtz said:You are wanting to show those two vectors are linearly independent. What about Hall's explanation don't you understand?
negation said:Hi Kurtz,
That wasn't what I wanted. I read my notes and it says that the basis is linearly independent.
Where did the zero vector came from?
LCKurtz said:But you don't know it is a basis until you show the vectors are linearly independent. And that is what you asked, as quoted in red above. Halls explained how you show that, and his explanation, including the definition of linear independence, shows why you set the linear combination equal to zero.