Do column 'vectors' need a basis?

R^2.In general, any vector can be written in terms of any basis, so long as the basis is a spanning set. Of course the coordinates of any one vector will be different in different bases, but the vectors themselves are the same.
  • #36
Mark44 said:
But here (-1) is shorthand for ##-1 \cdot (-\pi)##, so every representation of an element in the vector space ##\mathbb R## by its coordinate is implicitly in terms of the basis, ##-\pi##. That's been my point all along in this thread.

I think the point is that ##[-\pi]_{\beta} = (-1)##, that is the mapping to the coordinate vector w.r.t. ##\beta##. But the actual element of ##V## is ##-\pi##, a 1-tuple. The tuple itself, ##(-\pi)##, is not written in any basis, because it is the vector.
 
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  • #37
Mark44 said:
But here (-1) is shorthand for ##-1 \cdot (-\pi)##, so every representation of an element in the vector space ##\mathbb R## by its coordinate is implicitly in terms of the basis, ##-\pi##. That's been my point all along in this thread.

My point is that you must be able to identify the vectors in some basis-independent way. Otherwise, you can't define a basis in the first place, as you have no way to identify the specific basis vectors you are talking about.

Which I think was a point made by @SchroedingersLion
 
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  • #38
PeroK said:
My point is that you must be able to identify the vectors in some basis-independent way. Otherwise, you can't define a basis in the first place, as you have no way to identify the specific basis vectors you are talking about.
OK, that makes sense.
For me, the key to all of this is the realization that a vector space is "over" some field, the set from which the coordinates come. For a Euclidean space ##\mathbb R^n##, the field is the real numbers.
 
  • #39
To clarify some points. You can legally download Sheldon Axler : Linear Algebra Done Right 3rd edition. Refer to section on Linear Transformations. I think its 3.A from memory. Then go up 3:C (Matrix of a Linear Transformation with Respect to Bases).

This will clear up your misconceptions.
 
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  • #40
MidgetDwarf said:
To clarify some points. You can legally download Sheldon Axler : Linear Algebra Done Right 3rd edition. Refer to section on Linear Transformations. I think its 3.A from memory. Then go up 3:C (Matrix of a Linear Transformation with Respect to Bases).

This will clear up your misconceptions.

That's the one I'm currently working on, I managed to get it for free because of corona. It's very good but I haven't had the chance to read it through yet.

Thanks for the chapter, I'll have a look!
 
  • #41
etotheipi said:
That's the one I'm currently working on, I managed to get it for free because of corona. It's very good but I haven't had the chance to read it through yet.

Thanks for the chapter, I'll have a look!
The book is really good. There was one proof that through me off, it was more notational then anything, once I took Mathwonk advice to ditch the sum notation it was clear. In hindsight, it was a silly question. But yes, I think going through this will clear a lot of confusion you have in regards to a matrix of a linear transformation with respect to basis.

I had taken a linear algebra class out of Friedberg 4 years ago, but this was something I did not really understand. But Axler made it clear for me. I am currently on chapter 5 now.
 
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  • #42
etotheipi said:
I must have been reasoning along the lines of ##x^2 := (1,0,0)##, ##x := (0,1,0)##, ##1 := (0,0,1)##.

Which is not entirely false! The map ##(0,0,1) \mapsto 1, (0,1,0) \mapsto x, (1,0,0) \mapsto x^2## is a vector space isomorphism between ##\mathbb{R}^3## and the polynomials of degree at most ##2##.
 
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  • #43
When you say you have vector [itex](a,b,c)[/itex] in [itex]\mathbb R^3[/itex] that's usually meant w.r.t to the canonical basis. These numbers are coefficients of the linear combination of basis vectors corresponding to the given vector. When you change basis, but leave the coordinates the same, you get a different vector.
 

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