Basis BasisLet B = {b1, b2, ..., bn} be a basis of a vector space V. Then for any v ∈ V, we can find unique numbers c1, c2, ..., cn, such that
c1b1 + c2b2 + ... + cnbn = v.
The uniqueness of the coefficients make the following definition unambiguous.
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.