Basic properties and examples

Just as with coordinates on Rn, coordinates on general vector spaces mirror the sums and scalar multiples of the original vectors.

Suppose B is an ordered basis of a general vector space V.

For any vectors u and v in V,

[u + v]B = [u]B + [v]B

(the coordinates of a sum of vectors is the sum of the coordinates of the individual vectors)

For any vector v in V and any scalar c,

[cv]B = c[v]B

(the coordinates of a scalar multiple of a vector are the same scalar multiple of the coordinates of the vector)

For any vectors v1, v2, ..., vk in V and any scalars d1, d2, ..., dk,

[d1v1 + d2v2 + ... + dkvk]B = d1[v1]B + d2[v2]B + ... + dk[vk]B

(the coordinate vector of a linear combination of vectors in V is the same linear combination of their individual coordinate vectors).

The last of these three properties means that those relations among vectors in a general space which can be described in terms of linear combinations (e.g. spanning, linear independence and bases) translate into the same relations among the corresponding coordinate vectors in Rn. This gives a method of solving probelms in general, finite-dimensional spaces:

choose an appropriate basis of the space
represent all the vectors involved in terms of their coordinates with repsect to that basis
solve the problem for the coordinate vectors
translate the solution back into a solution to the original problem

Here are some sample problems.

Problem. Given the polynomials

p1(x) = 1
p2(x) = 2 + x
p3(x) = 3 + 2x + x2
p4(x) = 4 + 3x + 2x2 + x3

q(x) = –3 + 2x + x3,

show that {p1, p2, p3, p4} is a basis of P3 and write q in terms of this basis.

Solution. With respect to the standard basis E = {1, x, x2, x3} of P3, the vectors have coordinates

[p1]E = [1, 0, 0, 0]T,
[p2]E = [2, 1, 0, 0]T,
[p2]E = [3, 2, 1, 0]T,
[p2]E = [4, 3, 2, 1]T

and [q]E = [–3, 2, 0, 1]T.

Use the dependency algorithm on these coordinates. Stack them into a matrix

and reduce it to reduced row-echelon form:

.

Since the first four columns are have leading 1's, the first four columns of coordinates are linearly independent, so the polynomials p1, p2, p3, p4 are linearly independent. Since P3 has dimension 4, they form a basis of P3.

From the last column of the matrix,

[q]E = –7[p1]E + 3[p2]E – 2[p3]E + [p4]E

so (translating the coordinates back into the original vectors)

q(x) = –7p1(x) + 3p2(x) – 2p3(x) + p4(x).

Problem. Determine whether or not the matrices

form a basis of M22.

Solution. With respect to the standard basis E of M22, these matrices have coordinates

[A]E = [1, 3, 6, 7]T
[B]E = [0, 2, 4, 6]T
[C]E = [5, –1, –2, 4]T
[D]E = [–2, 0, 0, 3]T.

The determinant with these coordinates as columns is

Since row three is twice row two, this determinant is 0, so the coordinate columns do not form a basis of R4. Then the original matrices A, B, C, and D do not form a basis of M22.

If you're working with an inner product space and an orthonormal basis, you can even treat inner products of vectors as dot products of their coordinates.

If V is an inner product space with an ordered orthonormal basis B, then for any vectors u and v in V,

[u, v] = [u]B•[v]B

Proof. Suppose that B = {b1, b2, ..., bn}. Write the vectors u and v in in terms of B:

u = u1b1 + u2b2 + ... + unbn, so [u]B = [u1, u2, ..., un]T

v = v1b1 + v2b2 + ... + vnbn, so [v]B = [v1, v2, ..., vn]T

Then

[u, v] = [u1b1 + u2b2 + ... + unbn, v1b1 + v2b2 + ... + vnbn].

If you expand out the right-hand side, you get terms involving inner products of the vectors of B with each other. Since B is an orthonormal basis, [bi, bj] = 0 when i ≠ j. The only terms that "survive" the calculation are those with [bi, bi] in them for i = i, 2, ..., n:

[u, v] = u1v1[b1, b1] + u2v2[b2, b2] + ... + unvn[bn, bn].

Since the vectors b1, b2, ..., bn all have norm 1, [b1, b1] = [b2, b2] = ... = [bn, bn] = 1, so

[u, v] = u1v1 + u2v2+ ... + unvn = [u]B•[v]B.

Problem. Use coordinates to calculate the norm of the matrix

with respect to the inner product [A, B] = trace(ATB) on M23.

Solution. With respect to the standard basis E of M23, the matrix A has coordinates [A]E = [1, 2, 3, 3, 2, 1]T. Since the standard basis is orthonormal with respect to this inner product,

[A, A] = [A]E•[A]E = 12 + 22 + 32 + 32 + 22 + 12 = 28.

Then |A| = 281/2 = 2.71/2.