Can Extending Vectors from ℝm to ℝm+1 Alter Their Representation?

[daiviko]

So, I'm studying for my linear algebra midterm and I came up with kind of an interesting question that I pose to all of you brilliant people on physics forums.

Let's say you have a linear transformation T(x)=Ax, with A being an nxm matrice. Apparently, for this equation to hold, x must be a member of ℝ^m.

Maybe this is a ******** argument but if ℝ^m-1 is a subset/subspace (forgot the exact terminology) of ℝ^m then wouldn't the vector (2,1) in ℝ² be (2,1,0) in ℝ³? vector operations with vectors in ℝ^m (or at least as far as I know) can't create vectors in ℝ^m+1, right?

I have no idea if I really made my question clear at all, but I'm curious to hear what you guys have to say regardless.

 

[Jolb]

So, I'm studying for my linear algebra midterm and I came up with kind of an interesting question that I pose to all of you brilliant people on physics forums.

Let's say you have a linear transformation T(x)=Ax, with A being an nxm matrice. Apparently, for this equation to hold, x must be a member of ℝ^m.

Maybe this is a ******** argument but if ℝ^m-1 is a subset/subspace (forgot the exact terminology) of ℝ^m then wouldn't the vector (2,1) in ℝ² be (2,1,0) in ℝ³?

Not necessarily. What's to say the vector in ℝ³ isn't (0, 2, 1) or (2, 0, 1)? In fact, it could be infinitely many different things, even using the same basis. Saying that ℝ² is a subset of ℝ³ is analogous to saying that a given plane in a 3-D space is a subset of that 3-D space (granted it has to have some additional properties to be a subspace, like including zero). There are infinitely many different planes in ℝ³ that would qualify as a subspace, and any point in ℝ³ is inside a plane, so really (2,1) could represent infinitely many different points in ℝ³

vector operations with vectors in ℝ^m (or at least as far as I know) can't create vectors in ℝ^m+1, right? .

Yes they can! What if T(x) = Ax, where A is a matrix and x is a scalar? That goes from ℝ to a matrix space!