Coordinates of vector to the base

In summary, the basis vectors for base B can be found by multiplying the matrix of the given equations by the matrix whose columns are the basis vectors for base A. This will result in a matrix whose columns are the basis vectors for base B.
  • #1
twoflower
368
0
Hi all, I don't know how to solve this:

The vector u has the coordinates [itex]( u_1, u_2, u_3 ) [/itex] to the base [itex] a = { ( 1,0,0), (1,1,0), (1,1,1) }[/itex] and it has the coordinates [itex]( u^{'}_{1}, u^{'}_{2}, u^{'}_{3} ) [/itex] to the base b. We also know:

[tex]
u_1 = 2u^{'}_1 - u^{'}_2 + u^{'}_3
[/tex]

[tex]
u_2 = u^{'}_1 + 3u^{'}_2 - u^{'}_3
[/tex]

[tex]
u_3 = u^{'}_1 + u^{'}_2 - 4u^{'}_3
[/itex]

Find out the base b !

I have no idea how to solve it...I tried it but I couldn't find a way without adding unknown variables to the equations.

Could someone help please?
 
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  • #2
In base, A, u has the coordinates [itex]( u_1, u_2, u_3 )[/itex]. This means, explicitely, that u is

[tex]\vec{u}=u_1(1,0,0)+u_2(1,1,0)+u_3(1,1,1) \ \ (*)[/tex]

In base, B, u has the coordinates [itex]( u'_1, u'_2, u'_3 )[/itex]. This means, explicitely, that u is

[tex]\vec{u}=u'_1 \vec{b_1}+u'_2\vec{b_2}+u'_3\vec{b_3} \ \ (**)[/tex]

where the [itex]\vec{b_i}[/itex] are the vectors of base B, which is what we want to find.

Using the relations between the coordinates of base A and B, we can rewrite (*) has

[tex]\vec{u}=(2u'_1-u'_2+u'_3)(1,0,0)+(u'_1+3u'_2-u'_3)(1,1,0)+(u'_1+u'_2-4u'_3)(1,1,1)[/tex]

Rearanging,

[tex]\vec{u}=u'_1(4,2,1)+u'_2(...)+u'_3(...)[/tex]

So according to (**),

[tex]\vec{b_1}=(4,2,1), \vec{b_2}=(...), \vec{b_3}=(...)[/tex]
 
  • #3
the basis vectors b1, b2, and b3, are exactly the vectors that have coordinates
(u1', u2',u3') = (1,0,0), (0,1,0) and (0,0,1) in the basis B.

So to find the first vector say b1, all you have to do is plug in (u1', u2',u3') = (1,0,0), into those equations you are given, find u1, u2, and u3, and multiply those by the basis vectors in A.


I.e. you get u1 = 2, u2 = 1, u3 = 1, which means that b1 = 2(1,0,0) + 1(1,1,0) + 1(1,1,1) = (4,2,1), etc,,,, i.e. do the same thing to find b2, then b3.

computational questions like this are not what i love about linear algebra. linear algebra is a beautiful subject. this kind of thing has about as much to do with that subject as multiplying big numbers has to do with number theory. of course it is useful in a pinch to be able to calculate.

Lets call the original copy of R^n simply V. Then giving a basis is simply giving an isomorphism to another copy of R^n, that assigns differfent coordinates to each vector. For example, the basis A assigns the coordinates (1,0,0), (0,1,0), and (0,0,1) to the three vectors in A. I.e. it is given by a map V-->R^n that takes (1,0,0) to (1,0,0), and takes (1,1,0) to (0,1,0), and takes (1,1,1) to (0,0,1). It is not clear what the matrix of this map is.

To know the matrix of a map we need to know what it does to the standard basis. Thus writing down the matrix made from the basis A itself, i.e. with (1,0,0) in the first column, (1,1,0) in the second, and (1,1,1) in the third defines instead the map, R^n-->V, inverse to the coordinate system defined by A.

Similarly, the matrix whose columns contain the desired basis for B would define the inverse map R^n-->V of the coordinate system defined by B. If we can find this matrix its columns will solve our probloem.

Now what have we been given in the equations expressing u in terms of u'?

Well those equations give the A coordinates as a function of the B coordinates. So this is the map R^n-->V-->R^n composed of first the inverse of the B coordinate map, followed by the A coordinate system, i.e. by the B basis matrix followed by the inverse of the A - basis matrix.

i.e. the matrix of coefficients of the given equations is the matrix A^(-1)B where A is the matrix whose columns are the A basis, and where B is the matrix whose columns are the B basis. Since we have the A basis we have the matrix A, and we are also given the matrix A^(-1)B as the matrix of coefficients. Thus all we need to do to get the amtrix B is multiply

A by A^(-1)B. I.e. put the vectors of the basis A into the columns of a matrix called A, and multiply it by the matrix of coefficients of those equations expressing the u's in terms of the u's.

I realize this is incomprehensible, but it is clear if one can draw pictures with arrows in them showing the direction of the maps.
 
Last edited:

1. What are vector coordinates?

Vector coordinates refer to the set of numbers used to represent the position of a vector in space. They typically include an x-coordinate, a y-coordinate, and possibly a z-coordinate.

2. What is the base point of a vector?

The base point of a vector is the starting point or origin from which the vector is measured. It is often denoted by the coordinates (0,0,0).

3. How do you determine the coordinates of a vector?

To determine the coordinates of a vector, you need to know the magnitude and direction of the vector, as well as the coordinates of its base point. You can then use trigonometric functions to calculate the x, y, and z coordinates of the vector.

4. Can a vector have negative coordinates?

Yes, a vector can have negative coordinates. This simply means that the vector is pointing in the opposite direction of the positive coordinate axis. For example, a vector with coordinates (-3,2,1) would point 3 units to the left, 2 units up, and 1 unit forward from its base point.

5. How do you use vector coordinates in physics?

Vector coordinates are crucial in physics as they allow us to describe the position, velocity, and acceleration of objects in space. They are used in equations to calculate forces, energy, and other physical quantities.

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