The discussion focuses on proving that the set W = {x ∈ R^n | Ax = Bx} is a subspace of R^n, where A is an n*n matrix and B is a real number. Participants emphasize the need to demonstrate that W is closed under addition and scalar multiplication, as well as verifying that the zero vector is included in W. The solution involves showing that if x1 and x2 are in W, then x1 + x2 is also in W, and for any scalar a, ax is in W. The final proof confirms that W meets all criteria for being a subspace.
PREREQUISITESStudents of linear algebra, mathematicians, and anyone interested in understanding the properties of vector spaces and subspaces in R^n.
iamzzz said: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 please ..i have no idea what to do ...
Dick said:What's the definition of 'subspace'?
iamzzz said: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 .
Dick said: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?
iamzzz said: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?
Dick said:Sure it is. That means x1+x2 is in W, right? Now work on scalar multiplication.
Nitpick: It's a non-empty subset that that is closed under addition and scalar multiplication.iamzzz said:a subset of a vector space that is closed under addition and scalar multiplication
Fredrik said: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.
iamzzz said: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 ?
Dick said: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.
Fredrik said: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).