Vectors in R^{n} Definition and Properties We assume that you are already familiar with vectors in R^{2} and R^{3}, so that you will see that the definition extends naturally. A vector in R^{n} is a n x 1 matrix.
The set of all vectors in R^{n} is called nspace. We define the sum and difference of two vectors and the product of a scalar and a vector by just realizing that vectors are matrices.
Example Let then Properties of Vectors Vectors enjoy the following properties. If u, v, and w are vectors and c and d are scalars, then
The proofs of all of these come from the properties of real numbers since the definition of addition subtraction and scalar multiplication is given componentwise. For example to prove number 6, we have [(c + d)u]_{i} = (c + d)[u]_{i} = c[u]_{i} + d[u]_{i} = [cu]_{i} + [du]_{i} Dot Product and Length Just as we defined the dot product for vectors in R^{2} and R^{3}, we similarly define the dot product for two vectors in the more general R^{n}. Let
then the dot product (or scalar product) of u and v is defined by u ^{.} v = S u_{i}v_{i} We define the length or magnitude of a vector as
A vector is a unit vector if it has length one. The unit vector in the direction of u is given by
u
and the distance between u and v by Distance = u  v The angle q between two vectors is defined by
u . v u and v are orthogonal if u ^{.} v = 0
Example Let
as in the first example, then u ^{.} v = (1)(4) + (4)(2) + (3)(0) + (2)(4) = 12 and the distance between u and v is
The unit vector in the direction of u is
The angle between u and v is defined by
and
Properties of the Dot Product Next we give some basic properties of the dot product. Let u, v, and w be vectors and c be a scalar, then
We will prove property 2 and leave the rest of the proofs for you. Proof of 2 u ^{.} v = S u_{i}v_{i} = S v_{i}u_{i} = v ^{.} u Another important property called the CauchySchwartz inequality is important enough to stand by itself. CauchySchwartz Inequality If u and v are vectors then u ^{.} v < u v If equality holds, we say that u and v are parallel. This will occur only if u is a multiple of v.
Proof The proof is this is quite tricky. We let x be a scalar and note from property 1 that 0 < (xu + v) ^{.} (xu + v) = ^{ }u ^{.} u x^{2} + 2 u ^{.} v x + v ^{.} v Since the dot product produces a scalar, the above equation is a quadratic in x. A quadratic that is always positive has nonpositive discriminant. Hence (2 u ^{.} v)^{2}  4(u ^{.} u)(v ^{.} v) < 0 dividing by 4 gives (u ^{.} v)^{2}  (u ^{.} u)(v ^{.} v) < 0 or (u ^{.} v)^{2} < (u ^{.} u)(v ^{.} v) Taking the square root of both sides produces the result. The final property of the dot product that we will discuss comes from geometry. We all know that the shortest distance between two points is a straight line. If we want to travel from the origin to a point P, it is quicker to go along the straight path then to go to a point Q first and then to P. If we choose the point P to be the tip of the vector u + v, and Q to be the tip of the vector u, then we have a following diagram.
This leads us the the triangle inequality.
The Triangle Inequality If u and v are vectors then u + v < u + v
Proof We take the square of the left hand side u + v^{2} = (u + v) ^{.} (u + v) = u ^{.} u + 2 u ^{.} v + v ^{.} v = u^{2} + 2(u ^{.} v) + v^{2} < u^{2} + 2u v + v^{2} by CauchySchwartz = (u + v)^{2} Back to the Linear Algebra Home Page Back to the Math Department Home Page email Questions and Suggestions
