### Basis

##### 2. Coordinate

Let B = {b1, b2, ..., bn} be a basis of a vector space V. Then for any vV, we can find unique numbers c1, c2, ..., cn, such that

c1b1 + c2b2 + ... + cnbn = v.

The uniqueness of the coefficients make the following definition unambiguous.

The coordinates of v with respect to the basis B is the euclidean vector

[v]B = (c1, c2, ..., cn).

Example The formula

(x1, x2, x3) = x1(1, 0, 0) + x2(0, 1, 0) + x3(0, 0, 1) = x1e1 + x2e2 + x3e3

indicates that

[(x1, x2, x3)]{e1, e2, e3} = (x1, x2, x3).

In other words, when we say x1 is the first coordinate of the vector (x1, x2, x3), we really mean it is the first coordinate with regard to the standard basis {e1, e2, e3}. To illustrate the subtlety here, the formula

(x1, x2, x3) = x2e2 + x1e1 + x3e3

tells us

[(x1, x2, x3)]{e2, e1, e3} = (x2, x1, x3),

so that x1 is the second coordinate of (x1, x2, x3) with regard to the basis {e2, e1, e3}.

By the definition of the coordinates, we also have

[a0 + a1t + a2t2 + ... + antn]{1, t, t2, ..., tn} = (a0, a1, a2, ..., an).

In other words, the coefficients of a polynomial are the coordinates with regard to the standard monomial basis.

Finally, the equality

 [ a b ] = a[ 1 0 ] + b[ 0 1 ] + c[ 0 0 ] + d[ 0 0 ] = aE11 + bE12 + cE21 + dE22 c d 0 0 0 0 1 0 0 1

tells us

 [ a b ]{E11, E12, E21, E22} = (a, b, c, d). c d

Example In an earlier example, the system

 3x1 + x2 - x3 = 2 x1 - x2 + x3 = 2 2x1 + 2x2 + x3 = 6

was shown to have a unique solution x1 = 1, x2 = 1, x3 = 2. This means

1(3, 1, 2) + 1(1, -1, 2) + 2(-1, 1, 1) = (2, 2, 6),

Recall that in this case, the uniqueness of the solution actually implies that {(3, 1, 2), (1, -1, 2), (-1, 1, 1)} is a basis of R3. Moreover, the equality above means that the coordinates of (2, 2, 6) with regard to this basis is

[(2, 2, 6)]{(3, 1, 2), (1, -1, 2), (-1, 1, 1)} = (1, 1, 2).

In general, the coordinates of a vector (b1, b2, b3) with regard to the basis {(3, 1, 2), (1, -1, 2), (-1, 1, 1)} is obtained by solving the following system.

 3x1 + x2 - x3 = b1 x1 - x2 + x3 = b2 2x1 + 2x2 + x3 = b3

The solution is

 [ b1 ]{(3, 1, 2), (1, -1, 2), (-1, 1, 1)} = [ 3 1 -1 ]-1[ b1 ] b1 1 -1 1 b2 b3 2 2 1 b3 = [ 1/4 1/4 0 ] [ b1 ] -1/12 -5/12 1/3 b2 -1/3 1/3 1/3 b3 = [ (1/4)b1 + (1/4)b2 ]. - (1/12)b1 - (5/12)b2 + (1/3)b3 - (1/3)b1 + (1/3)b2 + (1/3)b3

In the example above, the coordinates of a vector v with respect to a basis of Rn is the unique solution of Bx = v, where the columns of B is the basis. Recall that columns being a basis is equivalent to the invertibility of B, and the solution of Bx = v is B-1v (for the invertible B). We then have the following general formula for coordinates with regard to any basis of Rn.

Let the columns of B = [b1 b2 ... bn] be a basis of Rn, then

[v]{b1, b2, ..., bn} = B-1v.