# Vectors and Co-vectors

1. Dec 8, 2007

### cathalcummins

Okay, So I have am elementary question to ask but it is of fundamental importance to me. First things first, I have been looking through the posts on "the difference between vectors and covectors'' and found them to be helpful. But not too conducive to the way I am trying to learn about them. The posts seem to revolve around tangent and cotangent spaces. Although I will eventually go on to use my definition of covectors and vectors to define natural bases on manifolds, I am trying to ascertain a stand alone'' version of the defintion of covectors and vectors.

I have begun with good ol' reliable $\mathbb{R}^3$ for my vector space: Let us define a vector space $V$ such that:

$$V=\mathbb{R}^3$$

$$V$$ is the set

$$V=\{ v:v^i e_i=(v^1,v^2,v^3)^T | v^i \in \mathbb{R}^3 \}$$

with basis, say

e_1=\left( \begin {array}{c} 1 \\ \noalign{\medskip} 0 \\ \noalign{\medskip} 0 \\ \end {array} \right)

e_2=\left( \begin {array}{c} 0 \\ \noalign{\medskip} 1 \\ \noalign{\medskip} 0 \\ \end {array} \right)

e_3=\left( \begin {array}{c} 0 \\ \noalign{\medskip} 0 \\ \noalign{\medskip} 1 \\ \end {array} \right),

To be more explicit, let me define what the vector is, say

$$v=(1,2,-5)^T$$

So that $v^1=1$, $v^2=2$ and $v^3=-5$. And so that:

$$v=1\cdot e_1+2\cdot e_2-5\cdot e_3$$

Now define $V^*$, a space dual to $V$, by its elements $f$;
$$V^*= \{ f: f=f_i e^i=(x,y,z) \}$$

so that $f_1=x$, $f_2=y$ and $f_3=z$.

with (covariant) basis:

$$e_1=\left( \begin {array}{ccc} 1 & 0 & 0 \\ \end {array} \right)$$

$$e_2=\left( \begin {array}{ccc} 0 & 1 & 0 \\ \end {array} \right)$$

$$e_3=\left( \begin {array}{ccc} 0 & 0 & 1 \\ \end {array} \right)$$

where it is demanded that

$$e^i(e_j)=\delta^i_j$$.

Further, if we want to know how the $f \in V^*$ acts on the $v \in V$, we must derive a relation:

$$f(v)=f(v^i e_i)=v^i f(e_i)=v^i \delta^j_i f(e_j)$$

But by our previous demand we have:

$$f(v)=v^i (e^j(e_i)) f(e_j)$$

By linearity we have:

$$f(v)=v^i (e^j(e_i)) f(e_j)$$

Now $v^i,f(e_j) \in \mathbb{R}$ so we can just shift them around at will.

$$f(v)=f(e_j) v^i (e^j(e_i))=f(e_j) v^i (e^j(e_i))$$

$$=(f(e_j) v^i e^j)(e_i)=(f(e_j) e^j)(v^i e_i)=(f(e_j) e^j)(v)$$

As this is true $\forall v \in V$ we must have:

$$} f \equiv f(e_j) e^j$$

For notational purposes we define $f_j=f(e_j)$. So that;
$$f \equiv f_j e^j$$

So back to the problem at hand:

$$f(v) = f_je^j(v)=f_1 e^1(v)+f_2 e^2(v)+f_3 e^3(v)$$

$$f(v) = f_je^j(1\cdot e_1+2\cdot e_2-5\cdot e_3)$$

$$=f_1 e^1(1\cdot e_1+2\cdot e_2-5\cdot e_3)+f_2 e^2(1\cdot e_1+2\cdot e_2-5\cdot e_3)+f_3 e^3(1\cdot e_1+2\cdot e_2-5\cdot e_3)$$

$$=f_1 e^1(1\cdot e_1)+f_2 e^2(2\cdot e_2)+f_3 e^3(-5\cdot e_3)$$

$=f_1 (1)+f_2 (2 )+f_3 (-5)$

And so

$f(v)=x (1)+y (2 )+z (-5)=x+2y-5z$

So $$f(v)$$ is a plane.

Right so my questions are:

1. What does $f(v)$ being a plane mean?

2. I know that $f(e_j)=f_j$ is just notation, and that it's form may be deduced from the given expression for $f$ and the fact that the bases of $V$ and $V^*$ abide $e^i(e_j)=\delta^i_j$ but what does $f(e_j)$ mean? Is it just $f$ acting on the basis elements of $V$? I mean, If i try to work out what $f(e_k)$ is from $$f(v)=f(e_j)e^j(v)[/itex], we just get a cyclic definition $f(e_k)=f(e_j)e^j(e_k)=f(e_j)\delta^k_j=f(e_k)$. And if so, I am finding it hard to define, say $f(e_3)\equiv f_3=z$. I mean would this be a valid description: $f(e_i)$ is "all of $f$" acting on the i-th basis component of the corresponding vector space. It is defined by producing the i-th component of the covector $f$ If anyone can clarify I'd be ever so grateful. 3. The form of $f$ I chose, relates to some sort of cartesian projection I think. Could someone shed some light on the situation. Cheers, edit: adjusted as requested. Last edited: Dec 9, 2007 2. Dec 8, 2007 ### Chris Hillman As a courtesy to other PF readers, I'd encourage you to learn how to take advantage of the latex markup feature in VB, the software used in PF; see this primer. After reading a few lines from that page, you should have no trouble reformatting your document. 3. Dec 8, 2007 ### Hurkyl Staff Emeritus If V is a vector space over R, then it's dual space $V^*$ is, by definition, the space of all linear transformations from V to R. (also called linear functionals) If we are studying a particular vector space, then it is conventional to call an element of its dual space a "covector". If we choose a basis for V, then there is a natural choice of basis for $V^*$; we call it the dual basis. And, generally, we write the coordinates of a vector as a column array and the coordinates of a covector as a row array. The reason is as follows: Let $v \in V$ and $\omega \in V^*$. Let $[v]$ be the coordinate representation of v with respect to our chosen basis, and similarly for $[\omega]$. Then, we have: [tex][\omega(v)] = [\omega][v]$$

(The left hand side is the 1x1 matrix containing the number $\omega(v)$. The right hand side is the product from matrix algebra, which yields a 1x1 matrix)

Last edited: Dec 8, 2007
4. Dec 8, 2007

### cathalcummins

>> If V is a vector space over R, then it's dual space $V^*$ is, by definition, the space of all linear transformations from V to R. (also called linear functionals)

Okay

>> If we are studying a particular vector space, then it is conventional to call an element of its dual space a "covector".

Okay

>> If we choose a basis for V, then there is a natural choice of basis for $V^*$; we call it the dual basis. And, generally, we write the coordinates of a vector as a column array and the coordinates of a covector as a row array.

Uh huh

>> The reason is as follows:

Let $v \in V$ and $\omega \in V^*$.
Let $[v]$ be the coordinate representation of v with respect to our chosen basis, and similarly for $[\omega]$.

Then, we have:
$$[\omega(v)] = [\omega][v]$$

(The left hand side is the 1x1 matrix containing the number $\omega(v)$. The right hand side is the product from matrix algebra, which yields a 1x1 matrix)

Okay

All that makes sense. It has given me some strength to know I am on the right track. However, my questions are still open, as far as my mind can stretch presently.

Last edited: Dec 8, 2007
5. Dec 23, 2007

### clarkyeah

For your first question, i give you an intutive sense of what 'the plane' means.
Assuming we are unknown of what shape the earth has. But by some experiments, we derive a function f(v), which return the potential energy for an object, where v could be any point on earth. Then f(v)=0 is where all points with zero potential energy. As you obtain that f(v)=0 is a plane, we may say the earth is flat.
The above explanation may not be strict. If i made any mistakes, please tell me.

6. Dec 25, 2007

### OrderOfThings

Not just arrows sitting at an origin are examples of vectors. There are other objects that also form vector spaces. Take a set of equally spaced lines in the plane. Enumerate the lines.

Can you think of what it means to:

• Multiply such a set of lines with a scalar? (easy)

• Add two such sets of lines together? (slightly more complicated)
What does the resulting set of lines look like in each case?

Also, think a little about linear coordinate systems. Not in the way of basis vectors, but rather in the form of a parallelogramic grid. What does it mean to read off the coordinates of a vector from such a grid?

Last edited: Dec 25, 2007