Hello,
I need some help understanding the concept of an arrow as a vector tensor.
A [a,b] tensor can be thought an arrow starting in (0,0) and ending in (a,b).
Can "arrows" begin somewhere else than (0,0)? If so, how does its "1 x n" notation look like?

10 hours ago, Magnificus58 said:

Hello,
I need some help understanding the concept of an arrow as a vector tensor.
A [a,b] tensor can be thought an arrow starting in (0,0) and ending in (a,b).
Can "arrows" begin somewhere else than (0,0)? If so, how does its "1 x n" notation look like?

First I would caution against referring to a "normal" vector as a tensor. Yes it is true that tensors are elements in a vector space, and therefore vectors, but I doubt that's what you mean. It is also true that, when talking about tensors in general, it is customary to refer to vectors as type 1,0 tensors but that is a harmless notational abuse which I advise you not to adopt for now.

To answer your question: yes, in the naive view of vecyors as arrows in a plane, they can have "heads" or "tails" anywhere, but to understand why you are never told this would require you to understand equivalence classes and stuff like that. Why bother?

Why not just use the grown-up definition of a vector? Namely a vector space comprises a set called V equipped with a closed binary operation called addition (closed just means that x+y is in V whenever x and y are), together with an associated scalar field which acts by multiplication on elements of the space, this also being a closed operation.

When feeling pompous, we talk of a vector space [i]over[/i] a certain field, and may even write it as V(F) (thankfully this is rare)

Don't even think about tensors until you are comfortable with all there is to know about vectors - there is quite a lot more

Edited September 6Sep 6 by Xerxes

6 hours ago, Magnificus58 said:

Hello,
I need some help understanding the concept of an arrow as a vector tensor.
A [a,b] tensor can be thought an arrow starting in (0,0) and ending in (a,b).
Can "arrows" begin somewhere else than (0,0)? If so, how does its "1 x n" notation look like?

I am not sure why you are introducing tensors, when you still have questions about vectors.

So I suggest we defer the introduction of tensors till we have dealt with your questions about vectors.

It would be helpful to tell us in what connection you wish to use vectors and tensors and if you understand matrices ?

OK so your arrow is what is also called a 'directed line segment'

I have shown this in Fig 1 as the arrow Op, where 9a,b) are the coordinates of p and V is the length of Op

Op is a line segment of the extended line shown dashed.

This line makes and angle theta with the x axis.

As you say this vector is one of a family of vectors that all start at the origin and are totally determined by their length v = Op and the anticlockwise angle theta they make with the x axis.

In other words they are 'bound' to one point (the origin) and indeed are called bound vectors.

7 hours ago, Magnificus58 said:

Can "arrows" begin somewhere else than (0,0)? If so, how does its "1 x n" notation look like?

Yes indeed if these vectors begin somewhere else they are called unbound vectors, as vector W in Fig 2

Now I am going to leave x y coordinates behind for a moment since it is obvious that an unbound vector will, in general, point to some other coordinate than (a, b) - let us choose one and call this point q again as in Fig2.

Now to specify a vector in a plane we need two numbers and if we do not want to be stuck with starting at zero we need to use different properties of the vector than its end coordinates.

So I am going to introduce the vector V with length v = Op and making an angle ϴ with the x axis as V = (v, ϴ)

Similarly we can consider the unbound vector W with length w = O'q and making an angle Φ with the x axis. So W = (w, Φ)

So if v = w

and simultaneously

ϴ = Φ

Then we have an unbound vector we can place anywhere on the plane parallel to our bound vector.

Does this help with your first question ?

Well, the terms "bound vector" and "unbound vector" are most certainly non-standard in mathematics. Maybe you could explain what you mean.

Also, recall that the "length" of a vector, and the angle between two vectors requires an inner product. And a vector space equipped with an inner product is called a metric space.

But not every vector space is a metric space. (Of course, those spaces whithout a metric are of no interest to physicists or enigineers!)

4 hours ago, Xerxes said:

Well, the terms "bound vector" and "unbound vector" are most certainly non-standard in mathematics. Maybe you could explain what you mean.

Also, recall that the "length" of a vector, and the angle between two vectors requires an inner product. And a vector space equipped with an inner product is called a metric space.

But not every vector space is a metric space. (Of course, those spaces whithout a metric are of no interest to physicists or enigineers!)

See

Fleisch

A students guide to Vectors and Tensors page 2

Cambridge University Press 2012

So if vectors are characterised by their magnitude and direction, dies that mean that two equally long vecotrs pointing in the same direction could in fact be considered to be the same vector ?
In other words if you were to move the vector shown inFig 1.1(a) to a different location, withour varying its lenght or pointing direction would it still be the same vector ?
In some applications the answer is yes and those vectors are called free vectors. You can move a free vector anywhere you like, as long as you don't change its length or direction, and it remains the same vector.
But in many Physics and Engineering problems you will be dealing with vectors that apply at a given location; such vectors are called bound or anchored vectors and you are not allowed to relocate bound vectors as you can free vectors.

You should also note that this thread is only about vectors that can be represented as directed line segments, or 'arrows', not any other type od vector.

Edited September 7Sep 7 by studiot

16 hours ago, studiot said:

See

Fleisch

A students guide to Vectors and Tensors page 2

Cambridge University Press 2012

You should also note that this thread is only about vectors that can be represented as directed line segments, or 'arrows', not any other type od vector.

So, it seems that both you and your source use the term "unbound vector" to refer to what most mathematicians would call a whole equivalence class of vectors. And since congruence in general is a broad (though not especially deep) subject, that is why I advised the opening poster not to think of vectors as directed line segments.

But since she, or he, seems to have lost interest in the subject, it's really not worth falling out over

4 minutes ago, Xerxes said:

So, it seems that both you and your source use the term "unbound vector" to refer to what most mathematicians would call a whole equivalence class of vectors. And since congruence in general is a broad (though not especially deep) subject, that is why I advised the opening poster not to think of vectors as directed line segments.

But since she, or he, seems to have lost interest in the subject, it's really not worth falling out over

Well the op certainly seems to have been frightened away, if that was your objective.

As far as I know most students start with the definition a vector is something that requires both a magnitude and a direction.

My treatment is linked to that and is designed to develop from there.

1 hour ago, studiot said:

Well the op certainly seems to have been frightened away, if that was your objective.

Of course it wasn't my objective, Why would you think such a thing? Surely you're not just trying to be offensive?

10 minutes ago, Xerxes said:

Of course it wasn't my objective, Why would you think such a thing? Surely you're not just trying to be offensive?

If you want to discuss your claim, then discuss it.

But, when presented with a counterexample, please do not pretend you are the sole arbiter of nomenclature in such an offensive way.