David Pierce | Matematik | M.S.G.S.Ü.

# Beyond three dimensions

## n-space

Let R be the set of real numbers (on which are defined the usual operations of multiplication, addition and additive inversion). Let Rn be the set of vectors (u1 u2 ... un)T. In the last section, we treated the geometry of R3. Properly understood, the same notions make sense in Rn for any n.

## Dot-product and norm

Suppose u and v are vectors in Rn. Their dot-product is given by the formula

u·v = u1v1 + u2v2 + ... + unvn.

In particular, u·u is never negative, so the norm of u can be defined to be the nonnegative scalar |u| such that

|u|2 = u·u.

Note in particular that u·u and |u| are positive if (and only if) u is not 0.

Call two vectors parallel if one is a scalar multiple of the other. (Note that 0 is a multiple of every vector.) Assume now that u is not 0. If u and v are parallel, then ku - v = 0 for some scalar k; otherwise, the equation is true for no scalar k. Considering the two cases separately yields the Cauchy–Schwarz Inequality:

|u·v| ≤ |u||v|,

with equality if and only if u and v are parallel.

Because of the Cauchy–Schwarz Inequality, there is a real number θ between 0 and 2π such that

u·v = |u||v| cos θ;

so you can think of θ as the angle between u and v. In particular, u and v are orthogonal when u·v = 0.

By doing the algebra, one finds

|u + v|2 = |u|2 + 2u·v + |v|2.

Applying Cauchy–Schwarz yields the triangle inequality:

|u + v| ≤ |u| + |v|.

One also has the Pythagorean Theorem: |u + v|2 = |u|2 + |v|2 when u and v are orthogonal.

Everything here makes sense geometrically, but it all follows from algebraic facts.

## An alternative definition

Instead of starting with the dot-product, we can define the norm so that

|u|2 = u12 + u22 + ... + un2.

Then we can define the dot-product by the equation

4u·v = |u + v|2 - |u - v|2.

## Linear transformations

### Theoretical definition

A linear transformation from Rn to Rm is a function T, such that T(x) is in Rm when x is in Rn, and satisfying the rules:

T(x + y) = T(x) + T(y),

T(kx) = kT(x).

Let ei be the vector in Rn which has 1 in row i and 0 everywhere else. If u is in Rn, then

u = u1e1 + u2e2 + ... + unen ,

and therefore

T(u) = u1T(e1) + u2T(e2) + ... + unT(en).

This will justify the following:

### Practical definition

A linear transformation from Rn to Rm is the function of multiplication by an m×n matrix; if A is such a matrix, then the corresponding linear transformation is denoted TA , and satisfies the rule:

TA(x) = Ax.

In particular, column i of A is just TA(ei). So if T is an arbitrary linear transformation, it is multiplication by the matrix

(TA(e1) TA(e2) ... TA(en));

we can denote this matrix by [T]. The equations

[TA] = A and T[T] = T

symbolize the equivalence of the two definitions of linear transformations.

### Linear operators

A linear transformation from Rn to itself is a linear operator on Rn. Suppose T is such. Its eigenvalues and eigenvectors are defined as for [T]. Suppose λ is an eigenvalue of T, with corresponding eigenvector u ; then u is not 0, and

T(u) = λu ;

so, T does not change the direction of u or its scalar multiples (although T collapses u to 0, if λ = 0). If [T] happens to be a diagonal matrix, then its diagonal entries are just the eigenvalues of T, and the vectors ei are eigenvectors. In this case, T effects a dilation or contraction or reflection or `collapse' in the direction of ei, depending on whether the corresponding eigenvalue is at least 1, between 1 and 0, negative, or 0.

A linear operator always has at least one eigenvalue, but the eigenvalues might not be real numbers. Such is the case in R2 if T effects rotation through some angle which is neither zero nor two right angles. If the angle is θ, then T(e1) = (cos θ sin θ)T and T(e2) = (-sin θ cos θ)T.

### Linear transformations as functions

Composition of linear transformations corresponds to multiplication of matrices:

TATB = TAB.

A linear transformation T is one-to-one if T(x) and T(y) are distinct whenever x and y are distinct. If T is one-to-one, and T is in fact a linear operator, then the matrix [T] is invertible, and T itself has an inverse, namely the linear operator T-1 such that

[T-1] = [T]-1.

Next section: Abstract vector spaces

Son değişiklik: Thursday, 14 January 2016, 15:38:41 EET