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.

If B = { v 1 , v 2 ,..., v m } is a basis for a subspace V , then any vector x in V can be written as a linear combination

in exactly one way.

Recall that to say B is a basis for V means that B spans V and B is linearly independent. Since B spans V , we can write any x in V as a linear combination of v 1 , v 2 ,..., v m . For uniqueness, suppose that we had two such expressions:

Subtracting the first equation from the second yields

Since B is linearly independent, the only solution to the above equation is the trivial solution: all the coefficients must be zero. It follows that c i − c A i for all i , which proves that c 1 = c A 1 , c 2 = c A 2 ,..., c m = c A m .

Consider the standard basis of R 3 from this example in Section 2.7:

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,

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.

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

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

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

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 .

Let

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”:

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

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

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

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

Let B = { v 1 , v 2 ,..., v m } be a basis of a subspace V . Finding the B -coordinates of a vector x means solving the vector equation

If x is not in V , then this equation has no solution, as x is not in V = Span { v 1 , v 2 ,..., v m } . In other words, the above equation is inconsistent when x is not in V .