solution01_wise2011

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Technische Universitt Mnchen WiSe 2011/12Fakultt fr Informatik Solution for Exercise Sheet 1

Dr. Tobias LasserRichard Brosig, Jakob Vogel

October 25, 2011

Exercises in Basic Mathematical Tools

Assignment 3 Cauchy-Schwary Inequality (Exemplary Solution)

Reminder Scalar Product

For two vectors x, yR n , dene the scalar product

x, y :=

n

i= 1 xi yi

with the following properties x, x , yR n , R :

a) Linearity

x + x , y = x, y + x , yx , y = x, y

b) Commutativity

x, y = y, x

c) Positive deniteness

x, x 0x, x = 0 x = 0R n

Proof Cauchy-Schwarz Inequality

x, y R n and R , it holds:

0 x + y, x + y = x, x + 2 x, y + 2 y, y (1)

Consider this equation in the special case y, y = 0, then:

y = 0

and thus0 x, x xR n

The Cauchy-Schwarz Inequality holds in this special case!

Therefore, let now w.l.o.g. 1 y, y > 0 and select

= x, yy, y

R .

1without loss of generality; general and well-used remark that a restriction can be made as the other cases havealready been dealt with


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Solution 1/ page 3

(iii) S := { f X | f ( 1) = 0} X is a subspace .

( f g) S :( f g)( 1) = f ( 1) g( 1) = 0

( f ) S :( f )( 1) = f ( 1) = 0

(iv) S := { f X | f ( x) = f (1 x) xR } X is a subspace .

( f g) S :

( f g)( x) = f ( x) g( x) = f (1 x) g(1 x) = ( f g)(1 x)

( f ) S :( f )( x) = f ( x) = f (1 x) = ( f )(1 x)

Note that these equalities are directly derived from the denitions of the addition andscalar multiplication operations.

(v) S := { f X | f ( x3) = f ( x)5 xR } X is no subspace.

Dene the constant function f ( x) = 1,

then f S , but 2 f /S as (2 f )( x3) = 2 f ( x3) = 2 = ( 2 f )( x)5 = ( 2 f ( x)) 5 = 25 .

(vi) S := { f X | f is continuous } X is a subspace .

Both, addition and scalar multiplication, are continuous functions themselves, and ap-plying them to further continuous functions does not change this property.Two mappings f , gS are continuous by denition of the set, and therefore ( f g)S and ( f ) S . These concepts will be discussed in detail in the lectures calculus part!

Assignment 5 Linear Dependence (Exemplary Solution)

In general, the question (for whether a linear combination exists) yields a system of linear equa-tions 2:

vi = xi + yi + zi i = 1, 2, 3

For the rst set x = ( 1, 1, 0) , y = ( 1, 0, 1) , z = ( 0, 1, 0) , one gets a particularly simple system:

3 = + 0 = + 2 =

From these equations, it is simple to solve for and :

= 3 = 1 = = 1

It is thus possible to express v as linear combination of these x, y, z.

2also called linear system


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Solution 1/ page 4

It is advisable to change (even standardize) the notation to matrix-vector-form. Then, the generalsystem is:

x1 y1 z1 x2 y2 z2 x3 y3 z3

= : A

= : x=

v1v2v3

= : bThe names A, x (unknowns), and b (right-hand side) are commonly used to refer to the threecomponents of the system, and there are many general (and specialized) approaches to solvingsuch problems. Some are going to be discussed in the lecture.

For the rst set, the matrix form is:

1 1 01 0 10 1 0

=302

1 1 0 31 0 1 00 1 0 2

The best-known solution algorithm is probably Gaussian Elimination which attempts to bring thesystem into triangular or even diagonal shape by adding up multiples of single rows, commutingrows, etc. Once that simple shape is reached, it is easy to get the solution.

a)1 1 0 31 0 1 00 1 0 2

I II I 1 0 0 11 0 1 00 1 0 2

II I 1 0 0 10 0 1 10 1 0 2

II II I 1 0 0 10 1 0 20 0 1 1

The three vectors x (red), y (green) and z (blue) are not linearly dependent themselves (i.e.in general conguration) and form a basis of R 3 , however, neither an orthogonal nor anorthonormal one.

b)1 2 1 31 3 0 00 1 1 2

II I 1 2 1 30 1 1 30 1 1 2

II + II I 1 2 1 30 0 0 10 1 1 2

In the second row of the system, a problematic situation has occurred. That part is logically

equivalent to the statement 0 = 1 which is always wrong. Consequently, the system doesnot have a solution.The three vectors x (red), y (green) and z (blue) turn out to be linearly dependent. Theyare no basis and do not generate R 3 , but just a two-dimensional plane within the three-dimensional space. Such matrices are called singular and said to be rank-decient .


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Solution 1/ page 5

The right-hand-side (black) points to a location outside of that plane, a solution is thereforeimpossible to exist. The three vectors just miss the information of the error vector (red,solid).

c)1 0 1 30 0 0 00 1 1 2

This time, the matrix is also singular , but the right-hand side vector (black) happens topoint into the plane generated by the selection of vectors x (red), y (green) and z (blue).

It is thus possible to express the right-hand side as linear combination of the three vectors,however, the solution is not unique. The reason for this phenomenon is that the third vector

contains information already available through the rst two, and is thus redundant.One can consequently nd an innite set of solutions, for instance (2, 1, 1) (orange) or(4, 3, 1) (violet).

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