Economic

Euclidean Spaces as Vector Spaces

Dr Damien S. Eldridge

Australian National University

10 February 2021

D. S. Eldridge (ANU) Euclidean Vector Spaces 10 February 2021 1 / 43

Readings

Anton, H (1987), Elementary Linear Algebra (fifth edition), John Wiley and Sons, USA: Chapters 1-2 and 4-6 (pp. 1-98 and 143-328).

Basilevsky, A (1983), Applied matrix algebra in the statistical sciences, The 2005 Dover unabridged republication of the original 1983 North Holland version, Dover Publications, USA: Chapters 2-5 (pp. 46-271).

Simon, CP, and L Blume (1994), Mathematics for economists, WW Norton and Company, USA: Chapters 6-11 and 26-28 (pp. 107-250 and 719-799).

Sundaram, RK (1996), A first course in optimization theory, Cambridge University Press, USA: Chapter 1, and Appendices A, B, and C.

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What is a Vector Space? (Part 1)

A vector space is a linear algebraic structure. Suppose that V is an arbitrary set of objects that we shall call vectors, + is a binary operation known as vector addition, · is a binary operation known as scalar multiplication of a vector, and (F , +̂, ·̂) is a field of scalars. Sets of scalars that you might encounter when looking at vector spaces include Q, R and C. The triplet (V , +, ·) is a vector space on a field of scalars (F , +̂, ·̂) if the following twelve axioms are satisfied.

(VS1) (Vectors exist): V 6= ∅. (VS2) (Scalars exist): F 6= ∅. (VS3) (Closure under vector addition): If x ∈ V and y ∈ V , then x + y ∈ V . (VS4) (Commutativity of vector addition): If x ∈ V and y ∈ V , then x + y = y + x. (VS5) (Associativity of vector addition): If x ∈ V , y ∈ V and z ∈ V , then x + (y + z) = (x + y) + z.

Continued on next slide. D. S. Eldridge (ANU) Euclidean Vector Spaces 10 February 2021 3 / 43

What is a Vector Space? (Part 2)

Continued from previous slide.

(VS6) (Additive identity): There is some vector 0 ∈ V such that 0 + x = x + 0 = x for all x ∈ V . (VS7) (Additive inverse): For each x ∈ V , there is some vector (−x) ∈ V such that x + (−x) = (−x) + x = 0. (VS8) (Closure under scalar multiplication): If a ∈ F and x ∈ V , then a ·x ∈ V . (VS9) (Distributivity of scalar multiplication with vector addition): If a ∈ F , x ∈ V and y ∈ V , then a · (x + y) = a ·x + a ·y and (x + y) ·a = x ·a + y ·b. (VS10) (Distributivity of scalar addition with vector multiplication): If a ∈ F , b ∈ F and x ∈ V , then (a+̂b) ·x = a ·x + b ·x and x · (a+̂b) = x ·a + x ·b. (VS11) (Commutativity of scalar multiplication): If a ∈ F , b ∈ F and x ∈ V , then a · (b ·x) = (a·̂b) ·x. (VS12) (Multiplicative identity): There is some scalar 1 ∈ F such that 1 ·x = x · 1 = x for all x ∈ V .

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Examples of Vector Spaces

(Rn, +, ·) over the field of real numbers is a vector space when addition and multiplication over scalars are the standard addition and multiplication operations for real numbers, vector addition is defined by x + y = (x1 + y1, x2 + y2, · · · , xn + yn), and scalar multiplication of a vector is defined by a ·x = (ax1, ax2, · · · , axn). This vector space is known as Euclidean n-sapace.

The set of all (m×n) matrices over the field of real numbers with matrix addition and standard scalar multiplication of a matrix is a vector space. This space is sometimes denote by Mmn.

The set of real valued functions defined on the entire real line (f : R −→ R) over the field of real numbers is a vector space when vector addition is defined by (f + g)(x) = f (x) + g(x) and scalar multiplication of a vector is defined by (kf )(x) = kf (x).

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Vector Sub-Spaces

Suppose that (V , +, ·) is a vector space on the field of scalars (F , +̂, ·̂). W is said to be a sub-space of V if W ⊆ V and (W , +, ·) is a vector space on the field of scalars (F , +̂, ·̂) when + and · are the same operations that were used in (V , +, ·). Theorem: If (V , +, ·) is a vector space on the field of scalars (F , +̂, ·̂) and W ⊆ V , then W is a sub-space of V if and only if both of the following conditions hold.

(Closure under vector addition): If x ∈ V and y ∈ V , then x + y ∈ V . (Closure under scalar mutliplication): If a ∈ F and x ∈ V , then a ·x ∈ V .

This theorem works because many of the vector space axioms are directly inherited by W from V , while some of the others are implied by the above two conditions and the inherited axioms.

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Linear Combinations

Suppose that (V , +, ·) is a vector space on the field of scalars (F , +̂, ·̂). Let xi ∈ V and ki ∈ F for all i ∈{1, 2, · · · , r}. A vector y is called a linear combination of the vectors x1, x2, · · · , xr if it can be expressed in the form

y = k1x1 + k2x2 + · · ·+ krxr = n

∑ i=1

kixi ,

where k1, k2, · · · , kr are scalars.

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Examples of Linear Combinations

Example 1: (9, 2, 7) is a linear combination of the vectors (1, 0, 0), (0, 1, 0) and (0, 0, 1) because (9, 2, 7) = 9(1, 0, 0) + 2(0, 1, 0) + (−7)(0, 0, 1). Example 2: (9, 2, 7) is a linear combination of the vectors (9, 0, 7) and (0, 1, 0) because (9, 2, 7) = 1(9, 0, 7) + 2(0, 1, 0).

Example 3: (9, 2, 7) is a linear combination of the vectors (1, 2,−1) and (6, 4, 2) because (9, 2, 7) = (−3)(1, 2,−1) + 2(6, 4, 2) for any (k1, k2). (Show how to find these values for k1 and k2 on the whiteboard.)

Example 4: (4,−1, 8) is not a linear combination of the vectors (1, 2,−1) and (6, 4, 2) because (9, 2, 7) 6= k1(1, 2,−1) + k2(6, 4, 2) for any (k1, k2). (Show this on the whiteboard.)

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The Spanning of Spaces

Suppose that (V , +, ·) is a vector space on the field of scalars (F , +̂, ·̂) and W ⊆ V is a subspace of V . Note that it is possible that W might be the entire space V , but this is not required to be the case.

The collection of vectors {x1, x2, · · · , xr} is said to span W if every vector in W can be expressed as a linear combination of the vectors in {x1, x2, · · · , xr}.

In effect, this means that we can generate the entire set of vectors in W by taking linear combinations of the vectors in {x1, x2, · · · , xr}.

Theorem: If {x1, x2, · · · , xr}. is a collection of vectors from the vector space V , then:

The set W of all linear combinations of {x1, x2, · · · , xr} is a sub-space of V ; and W is the smallest sub-space of V that contains {x1, x2, · · · , xr} in the sense that any other subspace of V that contains {x1, x2, · · · , xr} must itself contain W .

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Spanning Examples

The collection of vectors {(1, 0), (0, 1)} span R2. Consider an arbitrary vector (a, b) ∈ R2. Note that (a, b) = a(1, 0) + b(0, 1).

The collection of vectors {(1, 0, 0), (0, 1, 0), (0, 0, 1)} span R3. Consider an arbitrary vector (a, b, c) ∈ R3. Note that (a, b, c) = a(1, 0, 0) + b(0, 1, 0) + c(0, 0, 1).

The collection of vectors {(1, 1, 2), (1, 0, 1), (2, 1, 3)} do not span R3.

Work through this example on the white-board.

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Linear Independence

Suppose that (V , +, ·) is a vector space on the field of scalars (F , +̂, ·̂). Let S = {x1, x2, · · · , xr} be a collection of vectors from V and {k1, k2, · · · , kr} be a collection of scalars from F . The vector equation

k1x1 + k2x2 + · · ·+ krxr = 0

has at least one solution, namely k1 = k2 = · · · = kr = 0. If this is the only solution to the above vector equation, then S is called a linearly independent set of vectors.

If there are other solutions to the above vector equation, then S is called a linearly dependent set of vectors.

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Examples of Linear Dependence and Independence

S = {(1, 0, 0), (0, 1, 0), (0, 0, 1)} is a linearly independent set of vectors.

S = {(2, 1), (1, 2)} is a linearly independent set of vectors. Show this on the white-board

S = {(2,−1, 0, 3), (1, 2, 5,−1), (7,−1, 5, 8)} is a linearly dependent set of vectors.

Because 3x1 + x2 + (−1)x3 = 0. S = {(1,−2, 3), (5, 6,−1), (3, 2, 1)} is a linearly dependent set of vectors.

Show this on the white-board.

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Some Useful Results About Linear Independence

A set S containing two or more vectors is linearly dependent if and only if at least one of the vectors in S can be expressed as a linear combination of the other vectors in S.

Some special cases of this result include the following two results.

If a set of vectors S contains the zero vector, then it is linearly dependent. A set S containing exactly two vectors is linearly dependent if and only if at least one of the vectors is a scalar multiple of the others.

A set S containing two or more vectors is linearly independent if and only if no vector in S can be expressed as a linear combination of the other vectors in S.

If S = {x1, x2, · · · , xr} be a collection of vectors from Rn and r > n, then S is linearly dependent.

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A basis for a vector space

Suppose that (V , +, ·) is a vector space on the field of scalars (F , +̂, ·̂) and the set S = {x1, x2, · · · , xn} is a collection of vectors from V .

The set S is called a basis for the vector space V if both:

S is a linearly independent set of vectors; and The vectors in S span V .

If S = {x1, x2, · · · , xn} is a basis for a vector space V , then every set of vectors from V that contains more than n vectors is linearly dependent.

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Some basis examples

S = {(1, 0, 0, · · · , 0, 0), (0, 1, 0, · · · , 0, 0), · · · , (0, 0, 0, · · · , 0, 1)} is a basis for Rn.

This is sometimes called the standard basis for Rn.

S = {(1, 2, 1), (2, 9, 0), (3, 3, 4)} is a basis for R3. Show this on the white-board.

The set S = {M1, M2, M3, M4} is a basis for the vector space M22 of

(2 × 2) matrices with real entries, where M1 = (

1 0 0 0

) ,

M2 =

( 0 1 0 0

) , M3 =

( 0 0 1 0

) and M4 =

( 0 0 0 1

) .

Show this on the white-board.

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The dimension of a vector space

A non-zero vector space V is called finite dimensional if it has a basis with a finite number of vectors.

Any two bases for a finite dimensional vector space have the same number of vectors. The dimension of a finite vector space V is defined to be the number of vectors in a basis for V .

A non-zero vector space V is called infinite dimensional if it does not have a basis with a finite number of vectors.

By convention, the zero vector space is considered to be finite dimensional despite the fact that it contains no linearly independent vectors and hence has no basis.

By convention, the zero vector space is considered to have dimension zero.

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Some useful results about bases and dimension

If S = {x1, x2, · · · , xn} is a set of n linearly independent vectors from an n-dimensional vector space V , then S is a basis for V .

If S = {x1, x2, · · · , xn} is a set of n linearly independent vectors that spans an n-dimensional vector space V , then S is a basis for V .

If S = {x1, x2, · · · , xr} is a set of r linearly independent vectors from an n-dimensional vector space V and r < n, then S can be enlarged to form a basis for V .

In other words, there are vectors xr+1, xr+2, · · · , xn such that {x1, x2, · · · , xr , xr+1, xr+2, · · · , xn} is a basis for V .

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Vectors Contained Within a Matrix Part 1

Consider an (m×n) matrix A of the form

A =

 

a11 a12 a13 · · · a1n a21 a22 a23 · · · a2n a31 a32 a33 · · · a3n

... ...

... . . .

... am1 am2 am3 · · · amn

  .

This matrix contains m row vectors and n column vectors.

The set of row vector contained in A is SR = {a1•, a2•, · · · , am•}, where ai• = (ai1, ai2, ai3, · · · , ain).

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Vectors Contained Within a Matrix Part 2

The set of column vectors contained in A is SC = {a•1, a•2, · · · , a•n}, where

a•j = A =

 

a1j a2j a3j ...

amj

  .

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An example

Consider a (2 × 3) matrix A of the form

A =

( 2 1 0 3 −1 4

) .

The set of row-vectors of A are {(2, 1, 0), (3,−1, 4)}. The set of column vectors of A are

{ (2, 3)T , (1,−1)T , (0, 4)T

} .

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An Econometric Application: The Design Matrix

Recall the classical linear regression model Y = X β + e.

The matrix X in this model is known as the design matrix.

X is an (n×k) matrix that consists of n observations on each of k independent variables.

Each of the n rows of X consist of a single observation on each of the k independent variables that was obtained from one sample point. The sample point might, for example, be a particular household.

Each of the k columns of X consist of n observations on one of the k independent variables. There is one observation on each of these variables for every sample point. For example, one of the columns might be household wealth at a particular point in time for each of the n households in the sample.

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Vector Spaces Associated with a Matrix Part 1

Consider an (m×n) matrix A of the form

A =

 

a11 a12 a13 · · · a1n a21 a22 a23 · · · a2n a31 a32 a33 · · · a3n

... ...

... . . .

... am1 am2 am3 · · · amn

  .

Clearly this matrix contains m row vectors and n column vectors.

The sub-space of Rn that is spanned by the set of row-vectors of A is called the row-space of A.

The sub-space of Rm that is spanned by the set of column vectors of A is called the column-space of A.

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Vector Spaces Associated with a Matrix Part 2

Note that, except for the way in which the vectors are written, the row vectors of A are the column vectors of AT , while the column vectors of A are the row vectors of AT .

As such, the row-space of A is the same as the column-space of AT , while the column-space of A is the same as the row-space of AT . This means that we can apply the following two theorems to AT if we are interested in the column-space of the matrix A.

Theorem: Elementary row operations do not change the row-space of a matrix.

Theorem: The set of non-zero row vectors in any row-echelon form of a matrix is a basis for that matrix.

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A Row-Space Example Part 1

Suppose that we want to find a basis for row-space of the following matrix:

A =

 

1 −2 0 0 3 2 −5 −3 −2 6 0 5 15 10 0 2 6 18 8 6

  .

This means that we want to find a basis for the space that is spanned by the set of vectors {v1, v2, v3, v4}, where v1 = (1,−2, 0, 0, 3), v2 = (2,−5,−3,−2, 6), v3 = (0, 5, 15, 10, 0) and v4 = (2, 6, 18, 8, 6).

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A Row-Space Example Part 2

We will show on the white-board that a row-echelon form of the matrix A is 

 1 −2 0 0 3 0 1 3 2 0 0 0 1 1 0 0 0 0 0 0

  .

Thus we know that a basis for the row-space of the matrix A is given by the set of vectors {w1, w2, w3}, where w1 = (1,−2, 0, 0, 3), w2 = (0, 1, 3, 2, 0) and w3 = (0, 0, 1, 1, 0).

Note that while there are four rows in A, the dimension of the row-space of A is only three. The reason for this is that there are only three linearly independent rows in A.

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A Column-Space Example Part 1

Suppose that we want to find a basis for column-space of the following matrix:

A =

  1 0 1 13 2 5 1

0 4 4 −4

  .

This means that we want to find a basis for the space that is spanned by the set of vectors {v1, v2, v3, v4}, where v1 = (1, 3, 0), v2 = (0, 2, 4), v3 = (1, 5, 4) and v4 = (1, 1,−4). Note that the transpose of A is given by

AT =

 

1 3 0 0 2 4 1 5 4 1 1 −4

  .

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A Column-Space Example Part 2

We will show on the white-board that a row-echelon form of the matrix AT is 

 1 3 0 0 1 2 0 0 0 0 0 0

  .

Thus we know that a basis for the column-space of the matrix A is given by the set of vectors {w1, w2}, where w1 = (1, 3, 0) and w2 = (0, 1, 2).

Note that while there are four columns in A, the dimension of the column-space of A is only two. The reason for this is that there are only two linearly independent columns in A.

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The Rank of a Matrix

The dimension of the row-space of a matrix is the number of vectors in a basis for that space.

The dimension of the column-space of a matrix is the number of vectors in a basis for that space.

Theorem: If A is any matrix, then the row-space of A and the column-space of A have the same dimension.

The rank of a matrix is equal to the dimension of the row-space (and hence dimension of the column-space) of that matrix.

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An Example of the Rank of a Matrix Part 1

Consider the following matrix:

A =

  1 0 1 13 2 5 1

0 4 4 −4

  .

We have already shown that the dimension of the column-space of this matrix is equal to two.

Thus we know that the rank of this matrix will be equal to two.

Let us now show that the dimension of the row-space of this matrix is also equal to two.

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An Example of the Rank of a Matrix Part 2

We will show on the white-board that a row-echelon form of the matrix A is 

 1 0 1 10 1 1 −1 0 0 0 0

  .

Thus we know that a basis for the column-space of the matrix A is given by the set of vectors {w1, w2}, where w1 = (1, 0, 1, 1) and w2 = (0, 1, 1,−1). Note that while there are three rows in A, the dimension of the row-space of A is only two. The reason for this is that there are only two linearly independent rows in A.

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A Useful Collection of Equivalent Results

If A is an (n×n) matrix, then the following statements are equivalent:

A is invertible; Ax = 0 has only the trivial solution x = 0 (where x is an (n× 1) vector); In is the reduced-row-echelon form of A (where In is the (n×n) identity matrix); Ax = b is consistent for every (n× 1) vector b; det(A) 6= 0; A has rank n; The row vectors of A are linearly independent; and The column vectors of A are linearly independent.

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Some Additional Useful Results

Consider a system of linear equations that can be represented as a matrix equation of the form Ax = b.

This system of linear equations is consistent if and only if b is in the column-space of A.

b is in the column-space of A if it can be generated as a linear combination of the vectors in a basis of A.

This system of linear equations is consistent if and only if the rank of the coefficient matriz A is the same as the rank of the augmented-row-matrix (A|b). If this system of linear equations is consistent, contains m equations in n unknowns, and the rank of the coefficient matrix A is r, then the solution to this system of equations contains (n− r) parameters.

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Inner Products for Real Vector Spaces

An inner product on a real vector space V is a function 〈·,·〉 : V ×V −→ R that satisfies the following properties for all (u, v, w) ∈ V ×V ×V and all k ∈ R:

(IP1) (Symmetry): 〈u,v〉 = 〈v,u〉; (IP2) (Additivity): 〈u + v,w〉 = 〈u,w〉+ 〈v,w〉; (IP3) (Homogeneity): 〈ku,v〉 = k〈u,v〉; (IP4) (Non-Negativity): (a) 〈u,u〉 > 0 always, and (b) 〈u,u〉 = 0 if and only if u = 0.

A real vector space together an inner product is known as a real inner product space.

Inner products can also be defined for complex vector spaces. However, we will not deal with complex vector spaces in this course.

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The Euclidean Inner Product

The Euclidean inner product on Rn, where n ∈ N, is the function 〈·,·〉 : Rn × Rn −→ R defined by

〈u,v〉 = u1v1 + u2v2 + · · ·+ unvn = n

∑ i=1

uivi .

If u and v are row vectors, then 〈u,v〉 = u ·vT . If u and v are column vectors, then 〈u,v〉 = uT ·v. Euclidean n-space together with the appropriate Euclidean inner product forms an inner product space.

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Norms for Real Inner Product Spaces

If V is an inner product space, then the norm of a vector u ∈ V is defined to be

‖u‖ = 〈u,u〉 1 2 .

This is a measure of the length of the vector u, where length is interpreted as the distance from the origin to the point defining u in V .

The Euclidean norm on Rn, where n ∈ N, is

‖u‖ = {

n

∑ i=1

uiui

}1 2

=

{ n

∑ i=1

u2i

}1 2

.

A real inner product space together with a norm is known as a normed real vector space.

Example: Euclidean n-space is an example of a normed real vector space.

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The Relationship Between Norms and Distances

Suppose that V is a normed real vector space.

The distance between two points u ∈ V and v ∈ V , which is denoted by d(u, v), can be defined in terms of the norm as d(u, v) = ‖v −u‖. In the case of Euclidean n-space, we have

d(u, v) = ‖v −u‖ = {

n

∑ i=1

(v −u)2i

}1 2

=

{ n

∑ i=1

(vi −ui)2 }1

2

.

Note that this is just the Euclidean distance metric for Rn. As such, we can conclude that Euclidean n-space has all the desirable algebraic properties of a vector space and all the desirable topological properties of a metric space.

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The Angle Between Two Vectors

Suppose that V is a normed real vector space, u ∈ V and v ∈ V . The angle between u and v is defined by the following equation

cos θ = 〈u,v〉 ‖u‖‖v‖

where θ ∈ [0, π] is measured in radians. Note that this equation is only defined if u 6= 0 and v 6= 0.

Example: If u = (4, 3, 1,−2) and v = (−2, 1, 2, 3), then ‖u‖ = √

30, ‖v‖ =

√ 18 and 〈u,v〉 = −9. Hence we have

cos θ = 〈u,v〉 ‖u‖‖v‖

= −9

√ 30 √

18 = −3

2 √

15 .

Thus θ = cos−1( −3 2 √ 15 ).

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Orthogonality

Vectors are said to be orthogonal if they are at right-angles to each other.

In other words, vectors are said to be orthogonal if the angle between them is equal to 90 degrees. In other words, vectors are said to be orthogonal if the angle between them is equal to π2 radians. Recall that cos π2 = 0.

If V is a normed real vector space, u ∈ V , v ∈ V , u 6= 0 and v 6= 0, then u and v are orthogonal to each other if 〈u,v〉 = 0.

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Orthogonality in Euclidean Spaces

Consider Euclidean n-space with u ∈ Rn, v ∈ Rn, u 6= 0 and v 6= 0. If u and v are row vectors, then they will be othogonal if uvT = 0.

If u and v are column vectors, then they will be orthogonal if uT v = 0.

If X is an (n×k) matrix and u is an (n× 1) column vector, then X and u are orthogonal if X T u = 0(n×1), where 0(n×1) is the (n× 1) null (or zero) vector.

In effect, the matrix X and the column vector u will be orthogonal if every column vector contained in X is orthogonal to u.

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Convex Combinations

Consider X ⊆ Rn for some n ∈ N. Let (x, y) ∈ X ×X and λ ∈ [0, 1]. The vector wλ = λx + (1 − λ)y is called a convex combination of the vector x and the vector y.

Note that if x 6= y and λ ∈ (0, 1), then wλ 6= x and wλ 6= y. If x 6= y, then the entire line joining x and y can be generated by varying λ over the [0, 1] interval.

Example: (1 4

, 3 4 ) is a convex combination of (1, 0) and (0, 1) because

(1 4

, 3 4 ) = (1

4 )(1, 0) + (3

4 )(0, 1).

D. S. Eldridge (ANU) Euclidean Vector Spaces 10 February 2021 40 / 43

Convex Sets 1

Consider X ⊆ Rn for some n ∈ N. Let (x, y) ∈ X ×X and λ ∈ [0, 1]. Clearly wλ = λx + (1 − λ)y ∈ Rn for all λ ∈ [0, 1]. However, there are sometimes cases where wλ /∈ X for at least one λ ∈ [0, 1]. A set is called a convex set if, for every (x, y) ∈ X ×X and λ ∈ [0, 1], we have λx + (1 − λ)y ∈ X . Illustrate some examples on the white-board.

D. S. Eldridge (ANU) Euclidean Vector Spaces 10 February 2021 41 / 43

Convex Sets 2

A set is called a strictly convex set if, for every (x, y) ∈ X ×X such that x 6= y and λ ∈ (0, 1), we have λx + (1 − λ)y ∈ interior(X).

Recall that the boundary of a set X is denoted by ∂X . The interior of a set X can be defined as X \ ∂X . Of course, our definition of the boundary of a set X involved the use of a metric space containing X . But this is not a problem for sets contained in Euclidean n-space because of the relationship between the Euclidean norm and the Euclidean metric.

Clearly every strictly convex set is convex. However, there are some convex sets that are not strictly convex.

Illustrate some examples on the white-board.

D. S. Eldridge (ANU) Euclidean Vector Spaces 10 February 2021 42 / 43

Economic Application: Convex and Strictly Convex Preferences

Let % be a rational weak preference relation on RL+.

The “weakly preferred than” set for a particular consumption bundle x ∈ RL+ under % is given by U+x =

{ y ∈ RL+ : y % x

} .

A weak preference relation is said to be convex if U+x is a convex set for all x ∈ RL+. A weak preference relation is said to be strictly convex if U+x is a strictly convex set for all x ∈ RL+. Note the following.

All strictly convex weak preference relations are also convex. However, there are some convex weak preference relations that are not strictly convex. The “strictly preferred than” set for a particular consumption bundle x ∈ RL+ under % is given by U++x =

{ y ∈ RL+ : y % x, x � y

} .

Note that interior(U+x ) = U ++ x .

Illustrate some examples on the white-board.

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