Maths7

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Elements of linear algebra
7.1 Vectors

It is well known, that there is a one-to-one correspondence between the elements of the real number set and the points of a straight line. It is also known, that we can make a similar one-to-one correspondence between the points of the plane and the elements of set R2 of ordered real number pairs and between the points of the 3-dimensional space and the elements of set R3 of ordered real number triples. We generalize and introduce the n-dimensional space as the set Rn containing all the ordered real n-tuples, that is, n-element real sequences.
xi R, i = 1, . . . , n components of x.

Denition 7.1 The elements of Rn will be written in the form x = (x1 , x2 , . . . , xn ) and are said to be vectors. The real numbers x1 , . . . , xn are the
If y = (y1 , y2 , . . . , yn ) is another vector in Rn then we dene the sum:
x + y = (x1 + y1 , x2 + y2 , . . . , xn + yn )

and if is a real number we also dene the scalar multiplication:
x = (x1 , x2 , . . . , xn ).

Thus x + y Rn and also x Rn . Since the addition of vectors is led back to the addition of their coordinates, which are real numbers, the following properties hold true.

Proposition 7.1 Let x, y and z be arbitrary vectors in Rn , then 1. x + y = y + x, (the addition is commutative,) 2. (x + y) + z = x + (y + z) (associative,) 3. 0 Rn : x Rn : 0 + x = x, (there exists zero vector,) 4. x Rn : (-x) Rn : x + (-y) = 0. (each element has an opposite.)

Remark: The zero vector in Rn is obviously the vector 0, all of whose components are zero, and the opposite of a vector is obtained if its each components is replaced with its opposite.
Proposition 7.2 If x, y Rn and , µ R, then 1. (x + y) = x + y, 2. ( + µ)x = x + µx, 3. ( · µ)x = (µx),
4. 1 · x = x.

Denition 7.2 Let V be a set and an addition be dened on V satisfying the 1 - 4 properties (7.1) and for any real number let a map : V - V be dened satisfying the 1 - 4 properties (7.2) of scalar multiplication. Then V is called a real vector space.
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7.1.

VECTORS

· The directed line segments in the plane with xed initial point form a real

vector space, with the addition according to the parallelogram rule and scalar multiplication dened as follows: multiply the length of the vector by the absolute value of the number and change its direction to the opposite if the number is negative. above.

· Rn is a real vector space with the addition and scalar multiplication dened · The set of all functions of the form f : X - R is a real vector space, where the addition of two functions f and g is dened by (f + g)(x) := f (x) + g(x) x X, and for any R, (f )(x) := · f (x) denes the scalar multiplication.

Denition 7.3 Let V be a vector space and M a non-empty subset of V. M is said to be a subspace of V if it is a vector space with the addition and scalar multiplication dened in V.
7.1.1 Linear combination, linearly independent vector set

Let x1 , . . . , xk be arbitrary vectors in Rn and 1 , . . . , k be real numbers. Then the vector
1 x1 + · · · + k xk

is said to be a linear combination of vectors x1 , . . . , xk .

Denition 7.4 The set of vectors x1 , . . . , xk is 1 x1 + · · · + k xk = 0 = 1 = . . . = k = 0.

said to be linearly independent if

The only way to obtain the zero vector as a linear combination of a linearly independent set of vectors is the trivial one, that is, if we choose all the scalar coecients equal zero.

Denition 7.5 The system of vectors v1 , . . . , vk is called linearly dependent if there are scalars 1 , . . . , k such that at least one of them i = 0, 1 i k, but 1 v1 +
· · · + k vk = 0.

The subsequent statements are easily veried:
· Any non-empty subset of a linearly independent set of vectors is linearly in-

dependent.

· A set of vectors containing the zero vector is linearly dependent. · Any one element set containing only a non-zero vector is linearly independent. · A system of vectors having two equal vectors is linearly dependent. · A system of vectors is linearly dependent if and only if one of its vector is the

linear combination of the other vectors in the system.

· If v = 1 x1 + · · · + k xk and {x1 , . . . , xk } is linearly independent, then the scalars 1 , . . . , k are unique.

If H is an arbitrary subset of a vector space V, then the set of all linear combinations of vectors in H is called the linear hull of H and denoted by lin(H). The linear hull is a subspace of V.

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7.1.

VECTORS

Denition 7.6 The rank of a system of vectors {x1 , . . . , xk } is the number r if it has an r element linearly independent subset, but any at least r + 1 element subsystem is linearly dependent. Example 7.1 Let a = (2, 3, 4, 1) and b = (-1, 0, 1, 2) be two vectors in R4 Determine the vector a+2b.

Solution:
a + 2b = = (-2, -1, 0)} (2, 3, 4, 1) + (2 · (-1), 2 · 0, 2 · 1, 2 · 2) = (2 - 2, 3 + 0, 4 + 2, 1 + 4) = (0, 3, 6, 5)

Example 7.2 Decide if the system of vectors {a = (1, 2, 3), b = (-1, 0, 1), c = is linearly independent or dependent.

Solution: We have to nd out if the equation a + b + c = 0 implies that each of the coecients , , are zero. Comparing the coordinates of both sides of the equation we obtain the system of linear equations:
- - 2 2 - 3 + = = = 0 0 0

From the second equation it follows that = 2 and from the third one we have that = -3 and the rst equation is a consequence of these. Thus for any real number the = , = -3, = 2 triple satises the system of equations, therefore the vector system {a = (1, 2, 3), b = (-1, 0, 1), c = (-2, 0, 2)} is linearly dependent.

Denition 7.7 A set of vectors {v1 , . . . , vn } is called the basis (or coordinate system) of the vector space V if · it is linearly independent,
·

its linear hull is V.

Example 7.3 Verify that the set of vectors {e1 , . . . , en }, where ei is the vector with all 0s except for a 1 in the ith component, is a basis of the vector space Rn .

Solution: 1) The linear combination 1 e1 +· · ·+n en = (1 , . . . , n ) = 0 implies that 1 = . . . = n = 0, that is, {e1 , . . . , en } is linearly independent. 2) If x = (x1 , . . . , xn ) is an arbitrary vector in Rn , then obviously x = x1 e1 + · · · + xn en , therefore lin(e1 , . . . , en ) = Rn . It is impotent to emphasize that there are many bases in the vector spaces, for instance, any n element linearly independent vector set in Rn is a basis.
Denition 7.8 The dimension of a vector space is m if it has an m element basis. Denition 7.9 Let V = {v1 , . . . , vk } be a basis of the vector space V and x V. Then the scalars in the linear combination x = 1 v1 + · · · + k vk are called the coordinates of the vector x. In particular, j is the coordinate of x with respect to the basis vector vj . The column vector
xV = 1

. . .

k

is said to be the coordinate vector of x with respect to the basis V.
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7.1.

VECTORS

The coordinate vector depends on the basis. The same vector has dierent coordinate vectors with respect to dierent bases.
7.1.2 Elementary basis transformation

The problem is to nd out how the coordinate vector is changing if the basis is changed. Let V = {v1 , . . . , vi , . . . , vk } be a basis of the vector space V and w V a nonzero vector, w = 1 v1 + · · · + i vi + · · · + k vk and i = 0. Then the basis vector
vi = - 1 k 1 v1 - · · · + w - · · · - vk , i i i

therefore if the vector x as a linear combination of basis vectors in V was
x = 1 v1 + · · · + i vi + · · · + k vk ,

then substituting vi the linear combination
x = 1 v1 + · · · + i vi + · · · + k vk = 1 k 1 vk ) + · · · + k vk = = 1 v1 + · · · + i (- v1 - · · · + w - · · · - i i i i i i = (1 - 1 )v1 + · · · + w + · · · + (k - k )vk i i i

is obtained. The change can be better followed in the subsequent tables:
v1 w 1 x 1 v1 - x 1 - 1 ii
i i

. . .

. . . . . .

. . . . . .

. . .

. . . . . .

vi

. . .

i k

i k

w

. . .

vk

vk

k - k ii

In the left heading of the tables the basis vectors and in the top heading the vectors are listed and below in the j th row of the tables the coordinate of the corresponding vector with respect to the j th basis vector. Replacing the ith basis vector by the vector w is denoted by the frame around the ith coordinate of w. This coordinate is called pivot element. Notice that the pivot element is nonzero, this makes possible to exchange the vi basis vector with the vector w. In the second table the coordinates of the x vector is shown with respect to the new basis V = {v1 , . . . , w, . . . vk }. The coordinate of x with respect to the new basis vector w is the quotient of its original coordinate and the pivot element, while any other coordinate is obtained if from the original coordinate we subtract the product of this quotient and the corresponding coordinate of w. Thus the coordinate vectors of the x vector are
xV = 1

i . . . k

. . .

1 - 1 ii



and

xV

=

i i

. . . . . .

k - k ii

with respect to the bases V and V respectively.

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7.2.

INNER PRODUCT IN

RN

Some applications

Using elementary basis transformation actally any problem arising in linear algebra can be solved, that is, the linear algebra problems are easily solvable in appropriate coordinate system. To illustrate this statement we provide some example.

Example 7.4 Decide if the system of vectors
{y1 = (1, -1, 2, 0), y2 = (-1, 2, -2, 3), y3 = (3, -4, 6, -3)}

in R4 is linearly independent or dependent! Solution: The set of of vectors E = {e1 , e2 , e3 , e4 } is a basis in R4 . The coordinate vectors of y1 = (1, -1, 2, 0), y2 = (-1, 2, -2, 3), y3 = (3, -4, 6, -3) with respect to the basis E are
y1 E 1 -1 = 2 , 0 y2 E -1 2 = -2 3

and y3 E

3 -4 = 6 -3

Using a sequence of elementary basis transformations we change the original E basis taking as many yi vectors as we can into the new basis. The calculation is shown by the subsequent tables:
y1 e1 e2 e3 e4 y2 y3

1
-1 2 0

y1 -1 3 2 -4 - e2 -2 6 e3 3 -3 e4

y2 -1

1
0 3

y3 3 y1 -1 - y2 e3 0 e4 -3

y3 2 -1 0 0

The last table shows, that the new basis is {y1 , y2 , e3 , e4 } and the y3 can be expressed as a linear combination of y1 and y2 , namely y3 = 2y1 - 1y2 . Therefore the system {y1 , y2 , y3 } is linearly dependent.

Example 7.5 Decide if the vector x = (1, 0, 3) is {a1 = (1, 1, -1), a2 = (2, 1, 2), a3 = (0, 1, -4)}.

in the linear hull of vector system

Solution: Again begin with the basis of unit vectors {e1 , e2 , e3 } of R3 because the coordinates of the vectors in R3 with respect to this basis is the same as their components. Now the calculations are:
e1 e2 e3 a1 a 2 a3 x a1 1 2 0 1 - e2 1 1 1 0 -1 2 -4 3 e3 a2 2 -1 4 a3 x 0 1 a - 1 a2 1 -1 e3 -4 4 a3 2 -1 0 x -1 1 0

From the last table it can be concluded that x = -1a1 +1a2 , that is x lin(a1 , a2 , a3 ) Notice that the subspace lin(a1 , a2 , a3 ) is already spanned by the vectors a1 and a2 as well, since a3 is linear combination of them, that is, {a1 , a2 } is a basis of lin(a1 , a2 , a3 ) and therefore it is 2-dimensional subspace of R3 .
7.2 Inner product in

Rn

To generalize the geometric concepts known in the plane or 3-dimensional space some new notion must be introduced. 5

7.2.

INNER PRODUCT IN

RN

Rn

Denition 7.10 The inner product of vectors a =(a1, ..., an ) and b =(b1 , ..., bn ) is the real number
a, b = a1 b1 + .... + an bn .

Proposition 7.3 (Properties of inner product)

1. a, b Rn : a, b = b, a ,

2. a, b, c Rn : a + b, c = a, c + b, c , 3. a, b Rn , R : a, b = a, b , 4. a, a 0 and a, a = 0, i a = 0. Example: Let a = (3, -6, 1) and b = (1, 0, 4) R3 . Find the inner product Solution: 2a - 3b =(3, -12, -10) and a + 2b =(5, -6, 9) thus their inner product
2a - 3b, a + 2b = 3 · 5 + (-12) · (-6) + (-10) · 9 = -3. 2a - 3b, a + 2b .

another solution can be obtained if we use the properties of the inner product:
2a - 3b, a + 2b = = 2 a, a + a, b - 6 b, b = 2 · 46 + 7 - 6 · 17 = -3,

where a, a = 32 + (-6)2 + 12 = 46, a, b = 3 · 1 + (-6) · 0 + 1 · 4 = 7 and
b, b = 12 + 02 + 42 = 17.

7.2.1

Norm

Using the concept of inner product the norm of a vector can be dened:
n

x Rn : x :=

x, x =
i=1

x2 . i

The norm of a vector is obviously the generalization of the absolute value of real numbers, or the length of vectors in the 2-dimensional Cartesian coordinate systems.

Denition 7.11 Properties of norm

1. x 0 and x = 0 i x = 0, 2. x = || · x , 3. x + y x + y .
The rst two properties are easily seen, to prove the third one we need the Cauchy-Schwarz-inequality:

Theorem 7.4 (Cauchy-Schwarz-inequality) For any pair x, y Rn
| x, y | x · y .

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7.2.

INNER PRODUCT IN

RN

Proof: For any real number
x - y, x - y = x
2

- 2 x, y + 2 y

2

0,

therefore the discriminant of the quadratic polynomial
D = 4( x, y )2 - 4 x
2

y

2

0.

After rearranging, simplifying the inequality, and then extracting the square root of both side the desired statement is obtained. Then the third property of norm is obtained from the above inequality, because the square of the left side:
x+y
2

= x + y, x + y = x

2

+ 2 x, y + y 2 ,

and the square of the right side:
( x + y )2 = x
7.2.2 Distance in
2

+ 2 x · y + y 2.

Rn
x, y Rn : d(x, y) := x - y .

Using the norm of vectors dened above a distance function can be introduced in Rn : From the properties of the norm of vectors the expected properties of the distance function immediately follows: 1. d(x, y) 0 and d(x, y) = 0, i x = y, 2. d(x, y) = d(y, x), 3. d(x, y) d(x, z) + d(z, y). From the Cauchy-Schwarz-inequality it is obtained that
-1 x, y 1. x · y

Therefore there is a unique angle [0, ], such that
cos = is the x, y . x · y

angle enclosed by the vectors x and y.

Introducing the inner product in Rn it became possible to dene distance of vectors and measuring the angle of vectors. Therefore the Euclidean geometry of the plane or space can be generalized. Therefore the vector space Rn with the inner product is said to be Euclidean n-space.

Euclidean spaces

Example 7.6 Find the area of triangle with vertices a = (1, 1-1, 1), b = (2, 1, -1, 0) and c = (1, 2, 1, 1).

Solution: The length of sides in the triangle are: d(a, b) = b - a = (1, 0, 0, 1) = 2,d(a, c) = c - a = (0, 1, 2, 0) = 5, d(b, c) = c - b = (1, 0, 0, 1) = 7. The angle at the vertex a is 90 ,
because b - a, c - a = 0, therefore the area of the triangle is A = 7
5 2.

7.2.

INNER PRODUCT IN

RN

onal if their inner product is zero.

Cartesian coordinate system

Two vectors x, y in Rn are said to be orthog-

Denition 7.12 The set of vectors {x1 , . . . , xk } is an orthonormal set, if xi , xj = 0 if i = j, 1 if i = j.
An orthonormal set is linearly independent. If any vector v Rn is the linear combination of the vectors of the orthonormal set {x1 , . . . , xn }, then {x1 , . . . , xn } is called orthonormal basis, or Cartesian coordinate system. The set {e1 , . . . , en }, is a Cartesian coordinate system of the Euclidean n-space.

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