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Section3.8Bases as Coordinate Systems

Objectives
  1. Learn to view a basis as a coordinate system on a subspace.
  2. Recipes: compute the B -coordinates of a vector, compute the usual coordinates of a vector from its B -coordinates.
  3. Picture: the B -coordinates of a vector using its location on a nonstandard coordinate grid.
  4. Vocabulary word: B -coordinates.

In this section, we interpret a basis of a subspace V as a coordinate system on V , and we learn how to write a vector in V in that coordinate system.

Example

Consider the standard basis of R 3 from this example in Section 3.7:

e 1 = E 100 F , e 2 = E 010 F , e 3 = E 001 F .

According to the above fact, every vector in R 3 can be written as a linear combination of e 1 , e 2 , e 3 , with unique coefficients. For example,

v = E 35 2 F = 3 E 100 F + 5 E 010 F 2 E 001 F = 3 e 1 + 5 e 2 2 e 3 .

In this case, the coordinates of v are exactly the coefficients of e 1 , e 2 , e 3 .

What exactly are coordinates, anyway? One way to think of coordinates is that they give directions for how to get to a certain point from the origin. In the above example, the linear combination 3 e 1 + 5 e 2 2 e 3 can be thought of as the following list of instructions: start at the origin, travel 3 units north, then travel 5 units east, then 2 units down.

Definition

Let B = { v 1 , v 2 ,..., v m } be a basis of a subspace V , and let

x = c 1 v 1 + c 2 v 2 + ··· + c m v m

be a vector in V . The coefficients c 1 , c 2 ,..., c m are the coordinates of x with respect to B . The B -coordinate vector of x is the vector

[ x ] B = GKKI c 1 c 2 ... c m HLLJ in R m .

If we change the basis, then we can still give instructions for how to get to the point ( 3,5, 2 ) , but the instructions will be different. Say for example we take the basis

v 1 = e 1 + e 2 = E 110 F , v 2 = e 2 = E 010 F , v 3 = e 3 = E 001 F .

We can write ( 3,5, 2 ) in this basis as 3 v 1 + 2 v 2 2 v 3 . In other words: start at the origin, travel northeast 3 times as far as v 1 , then 2 units east, then 2 units down. In this situation, we can say that 3 is the v 1 -coordinate of ( 3,5, 2 ) , 2 is the v 2 -coordinate of ( 3,5, 2 ) , and 2 is the v 3 -coordinate of ( 3,5, 2 ) .

The above definition gives a way of using R m to label the points of a subspace of dimension m : a point is simply labeled by its B -coordinate vector. For instance, if we choose a basis for a plane, we can label the points of that plane with the points of R 2 .

Example

Let

v 1 = E 2 11 F v 2 = E 10 1 F .

These form a basis B for a plane V = Span { v 1 , v 2 } in R 3 . We indicate the coordinate system defined by B by drawing lines parallel to the v 1 -axis” and v 2 -axis”:

u 1 u 2 u 3 u 4 v 1 v 2 V

We can see from the picture that the v 1 -coordinate of u 1 is equal to 1, as is the v 2 -coordinate, so [ u 1 ] B = A 11 B . Similarly, we have

[ u 2 ] B = M 1 1 2 N [ u 3 ] B = C 3 2 1 2 D [ u 4 ] B = M 0 3 2 N .
Figure8Left: the B -coordinates of a vector x . Right: the vector x . The violet grid on the right is a picture of the coordinate system defined by the basis B ; one set of lines measures the v 1 -coordinate, and the other set measures the v 2 -coordinate. Drag the heads of the vectors x and [ x ] B to understand the correspondence between x and its B -coordinate vector.
Recipes: B -coordinates

If B = { v 1 , v 2 ,..., v m } is a basis for a subspace V and x is in V , then

[ x ] B = GKKI c 1 c 2 ... c m HLLJ means x = c 1 v 1 + c 2 v 2 + ··· + c m v m .

Finding the B -coordinates of x means solving the vector equation

x = c 1 v 1 + c 2 v 2 + ··· + c m v m

in the unknowns c 1 , c 2 ,..., c m . This generally means row reducing the augmented matrix

E ||| | v 1 v 2 ··· v m x ||| | F .