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).
A subspace is a subset of a vector space that is itself a vector space. It must satisfy three conditions: closure under vector addition, closure under scalar multiplication, and contain the zero vector.
To prove that a set is a subspace, you must show that it satisfies the three conditions for a subspace: closure under vector addition, closure under scalar multiplication, and contain the zero vector. In the case of W= {x member of R^n | Ax=Bx}, you must show that for any two vectors x and y in W, their sum (x+y) is also in W, and for any scalar c, the vector cx is also in W. You must also show that the zero vector (0) is in W.
The condition Ax=Bx ensures that the set W contains all vectors x that satisfy the equation Ax=Bx. This means that all vectors in W will have the same image under the linear transformation A. This is important because it guarantees that the set W will have a consistent structure and will behave like a vector space.
Yes, an example of a subspace in R^n is the set of all solutions to the equation Ax=0, where A is an n x n matrix. This set, also known as the null space of A, is a subspace of R^n because it contains the zero vector, and is closed under vector addition and scalar multiplication.
Proving that W is a subspace of R^n relates to linear transformations because it ensures that the set of all vectors that satisfy the equation Ax=Bx will behave like a vector space. This is important because linear transformations map vectors from one vector space to another, and in order for the image of a vector to be considered a vector, it must satisfy the properties of a vector space.