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Show that the set W= {x member of R^n | Ax=Bx} is a subspace of R^n

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Homework Statement


Let A be an n*n matrix and let B be a real number, Nothing which properties of matrix multiplication you use, show that the set W= {x member of R^n | Ax=Bx} is a subspace of R^n, where x in the equation Ax=bx is represented by a column vector instead of an n-tuple with commas.


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The Attempt at a Solution



Can i have some hint plz ..i have no idea what to do ...

Homework Statement





Homework Equations





The Attempt at a Solution

 

Answers and Replies

  • #2
Dick
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Homework Statement


Let A be an n*n matrix and let B be a real number, Nothing which properties of matrix multiplication you use, show that the set W= {x member of R^n | Ax=Bx} is a subspace of R^n, where x in the equation Ax=bx is represented by a column vector instead of an n-tuple with commas.


Homework Equations





The Attempt at a Solution



Can i have some hint plz ..i have no idea what to do ...

Homework Statement





Homework Equations





The Attempt at a Solution

What's the definition of 'subspace'?
 
  • #3
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What's the definition of 'subspace'?
a subset of a vector space that is closed under addition and scalar multiplication
Let H be a homogeneous system of linear equations in x1, ...,xn . Then the subset S of R^n which consists of all solutions of the system H is a subspace of R^n .
 
  • #4
Dick
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a subset of a vector space that is closed under addition and scalar multiplication
Let H be a homogeneous system of linear equations in x1, ...,xn . Then the subset S of R^n which consists of all solutions of the system H is a subspace of R^n .
So can you show W is closed under addition and scalar multiplication? That's what they want. Suppose x1 and x2 are in W. Is x1+x2 in W?
 
  • #5
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So can you show W is closed under addition and scalar multiplication? That's what they want. Suppose x1 and x2 are in W. Is x1+x2 in W?
Yes Ax1=Bx1 =>add Ax2 on both side=> Ax1+Ax2=Bx1+Bx2 => A(x1+x2)=B(x1+x2) is this enough to show close under addition?
 
  • #6
Dick
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Yes Ax1=Bx1 =>add Ax2 on both side=> Ax1+Ax2=Bx1+Bx2 => A(x1+x2)=B(x1+x2) is this enough to show close under addition?
Sure it is. That means x1+x2 is in W, right? Now work on scalar multiplication.
 
  • #7
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Sure it is. That means x1+x2 is in W, right? Now work on scalar multiplication.
Ax1=Bx1 =>mutiply x2 on both side=> Ax1*x2=Bx1*x2 => A(x1*x2)=B(x1*x2) is this enough to show close under sclar multiplication ?
 
  • #8
Fredrik
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a subset of a vector space that is closed under addition and scalar multiplication
Nitpick: It's a non-empty subset that that is closed under addition and scalar multiplication.

An alternative (but equivalent) definition says that a subspace of a vector space V is a subset U that's also a vector space, with its addition and scalar multiplication operations defined as the restrictions of the original addition and scalar multiplication operations to U.

So in addition to the two closure properties, you should also verify that the subset is non-empty. Usually, the easiest way to do this is to verify that the 0 vector of the original vector space is a member of the subset.
 
  • #9
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Nitpick: It's a non-empty subset that that is closed under addition and scalar multiplication.

An alternative (but equivalent) definition says that a subspace of a vector space V is a subset U that's also a vector space, with its addition and scalar multiplication operations defined as the restrictions of the original addition and scalar multiplication operations to U.

So in addition to the two closure properties, you should also verify that the 0 vector of the original vector space is a member of the subset.
Thank you
 
  • #10
Dick
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Ax1=Bx1 =>mutiply x2 on both side=> Ax1*x2=Bx1*x2 => A(x1*x2)=B(x1*x2) is this enough to show close under sclar multiplication ?
Same answer, yes. So if x1 is in W then x1*x2 is. But I am a little worried if you know what both sides mean. A(x1)*x2 is A(x1*x2) because A is a matrix with x2 a scalar and x1 a vector. (B*x1)*x2=B*(x1*x2) because x1 is a vector and B and x2 are scalars. They are a little different. If you understand this, then no problem.
 
  • #11
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Same answer, yes. So if x1 is in W then x1*x2 is. But I am a little worried if you know what both sides mean. A(x1)*x2 is A(x1*x2) because A is a matrix with x2 a scalar and x1 a vector. (B*x1)*x2=B*(x1*x2) because x1 is a vector and B and x2 are scalars. They are a little different. If you understand this, then no problem.
Thank you for all the help.
 
  • #12
Fredrik
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Since you have completed the problem, I can show you how I would do it:

0 is in W since A0=0=B0.

For all a,b in ℝ, and all x,y in W, we have A(ax+by)=aAx+bAy=aBx+bBy=B(ax+by), so ax+by is in W. (It's of course fine to break this up into two parts, one for scalar multiplication and one for addition).
 
  • #13
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Since you have completed the problem, I can show you how I would do it:

0 is in W since A0=0=B0.

For all a,b in ℝ, and all x,y in W, we have A(ax+by)=aAx+bAy=aBx+bBy=B(ax+by), so ax+by is in W. (It's of course fine to break this up into two parts, one for scalar multiplication and one for addition).
Thanks I should add A0=B0
 

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