Change of basis in R^n and dimension is <n

IniquiTrance

Suppose I have a basis for a subspace V in [tex]\mathbb{R}^{4}[/tex]:

[tex]\mathbf{v_{1}}=[1, 3, 5, 7]^{T}[/tex]
[tex]\mathbf{v_{2}}=[2, 4, 6, 8]^{T}[/tex]
[tex]\mathbf{v_{3}}=[3, 3, 4, 4]^{T}[/tex]

V has dimension 3, but is in [tex]\mathbb{R}^{4}[/tex]. How would one switch basis for this subspace, when you can't use an invertible transition matrix?

Thanks!

hgfalling

I don't really understand the question very well. "Changing basis" is a process of converting the representation of vectors in a space from one basis to another. But here you have two different spaces. It's kind of like asking "how do you change basis between [itex] \mathbb{R}^{3} [/itex] and [itex] \mathbb{R}^{2} [/itex]?"...well, you can't.

IniquiTrance

Hmm, I guess I'm asking how one would switch to another another basis that spans V. Given co-ordinates in terms of v₁, v₂, v₃, how would one find co-ordinates with respect to a new spanning set of vectors?

If V spanned R⁴, then I could just use a transition matrix. It's the fact that V is dimension 3, but each basis vector has 4 co-ordinates, that is confusing me.

hgfalling

Well, if they gave you coordinates in terms of v₁, v₂, and v₃, then you'd have a vector with 3 components, right? And any other basis for the same space would also have 3 components, because the dimension of the subspace is 3. So then you have your usual 3x3 thing for changing basis between those.

IniquiTrance

But how can you use a square matrix when the co-ordinate vectors have 4 components? This is what's confusing me...

Office_Shredder

You don't have four components, because instead of writing out vectors of v in terms of the standard basis in R⁴, you write them out in coordinates in terms of v₁, v₂ and v₃. Then you have three coordinates as required. For example, the vector (1,3,1) in this coordinate system would be 1*v₁+3v₂+1v₃=(10,18,27,35)

IniquiTrance

FFFFFFFFFFFUUUUUUUUUUUUUUUUU-

Great explanation!!

Thanks for the responses!