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Linear Algebra Proof

  • #1

Homework Statement



SHow that the set of solutions to a homogenous system of m linear equations in n variabes is a subspace of [itex]ℝ^{n}[/itex] (Show that this set satisfies the definition of a subspace)

Homework Equations





The Attempt at a Solution


If {V1,...Vk}=[itex]ℝ^{n}[/itex] then every vector [itex]\vec{q}[/itex][itex]\in[/itex]ℝ can be written as a linear combination of the set
c1V1+....+ckVk=[itex]\vec{q}[/itex]
This system of linear equations must have a solution for every [itex]\vec{q}[/itex][itex]\in[/itex]ℝ and therefore the rank of the coefficient matrix = n
If the rank of the coefficient matrix of a system
c1V1+...+ckVk=v
is n, then the system is consistent for all V[itex]\in[/itex]ℝ
∴ {V1,....,Vk}=[itex]ℝ^{n}[/itex]


I thought I was on the right track, but a theorem in my textbook says
" Let [A|[itex]\vec{b}[/itex]] be a system of m linear equations in n variables. Then [A|[itex]\vec{b}[/itex]] is consistent for all [itex]\vec{b}[/itex]=[itex]ℝ^{n}[/itex] if and only if rank(A)=m"

Does the requirement change is they are homogenous? Am I even on the right track?
 

Answers and Replies

  • #2
Ray Vickson
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Homework Statement



SHow that the set of solutions to a homogenous system of m linear equations in n variabes is a subspace of [itex]ℝ^{n}[/itex] (Show that this set satisfies the definition of a subspace)

Homework Equations





The Attempt at a Solution


If {V1,...Vk}=[itex]ℝ^{n}[/itex] then every vector [itex]\vec{q}[/itex][itex]\in[/itex]ℝ can be written as a linear combination of the set
c1V1+....+ckVk=[itex]\vec{q}[/itex]
This system of linear equations must have a solution for every [itex]\vec{q}[/itex][itex]\in[/itex]ℝ and therefore the rank of the coefficient matrix = n
If the rank of the coefficient matrix of a system
c1V1+...+ckVk=v
is n, then the system is consistent for all V[itex]\in[/itex]ℝ
∴ {V1,....,Vk}=[itex]ℝ^{n}[/itex]


I thought I was on the right track, but a theorem in my textbook says
" Let [A|[itex]\vec{b}[/itex]] be a system of m linear equations in n variables. Then [A|[itex]\vec{b}[/itex]] is consistent for all [itex]\vec{b}[/itex]=[itex]ℝ^{n}[/itex] if and only if rank(A)=m"

Does the requirement change is they are homogenous? Am I even on the right track?
You are on the wrong track. You need to worry about the rank of the matrix. For example, what is the dimensionality of the solutions to the following linear system?
[tex] x_1 + x_2 + x_3+ x_4 = 0\\
2x_1 + 2x_2 + 2x_3 + 2x_4 = 0
[/tex]
 
  • #3
jbunniii
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I think you're making this much harder than it needs to be. You don't need to know the rank of the system.

A homogeneous system can be written as follows: ##Ax = 0##, where ##A## is some matrix.

A vector ##x## is a solution to the system if and only if ##Ax = 0##. So the set of solutions is precisely the set of vectors which satisfy the equation.

What are the requirements to show that a set is a subspace?
 
  • #4
You are on the wrong track. You need to worry about the rank of the matrix. For example, what is the dimensionality of the solutions to the following linear system?
[tex] x_1 + x_2 + x_3+ x_4 = 0\\
2x_1 + 2x_2 + 2x_3 + 2x_4 = 0
[/tex]
Not sure what you mean by dimentionality of the system
but the RREF form yields the system
[tex] x_1 + x_2 + x_3+ x_4 = 0
[/tex]
 
  • #5
I think you're making this much harder than it needs to be. You don't need to know the rank of the system.

A homogeneous system can be written as follows: ##Ax = 0##, where ##A## is some matrix.

A vector ##x## is a solution to the system if and only if ##Ax = 0##. So the set of solutions is precisely the set of vectors which satisfy the equation.

What are the requirements to show that a set is a subspace?
Set is a subspace if it is closed under addition and scalar multiplication (and therefore include the zero vector)
 
  • #6
Ray Vickson
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Not sure what you mean by dimentionality of the system
but the RREF form yields the system
[tex] x_1 + x_2 + x_3+ x_4 = 0
[/tex]
Yes, and that is my point exactly: the matrix has rank 1, so the dimensionality of the solution set is 4 - 1 = 3. This counters your statement in the OP that the rank of A is n.

However, as has already been pointed out, you don't even need to know this to do the question; you would need to know it if you were asked about the dimension of the subspace.
 
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  • #7
jbunniii
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Set is a subspace if it is closed under addition and scalar multiplication (and therefore include the zero vector)
OK, let's start with addition. If ##x## and ##y## are solutions, what can you say about ##x+y##?
 
  • #8
OK, let's start with addition. If ##x## and ##y## are solutions, what can you say about ##x+y##?
That x+y is also a solution? It should be if it's closed under addition
 
  • #9
jbunniii
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That x+y is also a solution? It should be if it's closed under addition
Are you asking me or telling me? You need ##A(x+y) = 0##. Is it true? If so, why?
 

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