Lecture 1 Vector Spaces

8/12/2019 Lecture 1 Vector Spaces

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September 24, 2010

LINEAR ALGEBRA

RODICA D. COSTIN

Let us measure everything that is measurable, andmake measurable everything that is not yet so.Galileo Galilei

Contents

1. Vector spaces 1

1.1. Notations 11.2. The definition of vector spaces 21.3. Subspaces 41.4. Linear Span 51.5. Linear independence 61.6. Basis and Dimension 6

1. Vector spaces

1.1. Notations.x Sdenotes the fact that the element x belongs to the set S.Zdenotes the set of all integer numbers,Z+ denotes the set of all positive integers,Rdenotes the set of real numbers,Cdenotes the set of complex numbers.

1


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2 RODICA D. COSTIN

Of course, R C(the set of real numbers is included in the set of complexnumbers).

While in planar geometry it is customary to denote the coordinates ofpoints by (x, y), in linear algebra it is often preferable to use (x1, x2).Similarly, instead of denoting generic coordinates in space by (x , y , z), it

may be preferable using (x1, x2, x3).In Linear Algebra the components of vectors are listed vertically, so I

should write them as (x1, x2)T. I will start using the correct notation as

soon as it matters, namely when they start being multiplied by matrices.

1.2. The definition of vector spaces.

1.2.1. Usual examples. Recall the vectors in the plane. You can think ofthem as geometric representations (arrows), or as points, with coordinates(x1, x2). The translation between the two ways of representing vectors

1

in the plane is: given a vector represented by an arrow, then (x1, x2) are thecoordinates of the endpoint of the arrow if its starting point is placed at theorigin. We denote by R2 the set of vectors in the plane:

R2 ={(x1, x2) |xj R}

Similarly, you can think of vectors in space as arrows, or as coordinates(x1, x2, x3), and R

3 denotes the set of vectors in space:

R3 ={(x1, x2, x3) |xj R}

You may have noticed that the set of planar vectors and the set of vec-tors in space are quite similar: they can added and subtracted, they canbe multiplied by scalars (by which we mean real numbers). Representingvectors in terms of coordinates, the operations are very simple: they aredone coordinate-by-coordinate. For example, addition of vectors in R2 is:

(x1, x2) + (y1, y2) = (x1+ y1, x2+ y2)

and multiplication by scalars is:

c(x1, x2) = (cx1, cx2) for c R

while in R3 the operations are

(x

1, x

2, x

3) + (y

1, y

2, y

3) = (x

1+y

1, x

2+y

2, x

3+y

3)c(x1, x2, x3) = (cx1, cx2, cx3) for c R

Similarly, we can define n-dimensional vectors by

Rn ={(x1, x2, . . . , xn) ; xj R}

1As a mathematical notion, this is a one-to-one and onto map, taking arrows into

coordinates.


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LINEAR ALGEBRA 3

which we can add by:

(x1, x2, . . . , xn) + (y1, y2, . . . , yn) = (x1+ y1, x2+ y2, . . . , xn+ yn)

and multiply by scalars:

c(x1, x2, . . . , xn) = (cx1, cx2, . . . , c xn) for c R

It turns out that there is a great advantage to allow for complex coordi-nates, in which case we consider C2:

C2 ={(z1, z2) ; z1,2 C}

which can be added:

(z1, z2) + (w1, w2) = (z1+ w1, z2+ w2)

and multiplied by scalars which are complex numbers:

c(z1, z2) = (cz1, cz2) for c C

Similarly, we can consider C3, . . .Cn.Here is the abstract definition which summarizes the properties of addition

and multiplication by scalars of the vector spaces listed above.

1.2.2. Abstract Definition. Let F = Ror C.

Definition 1. A vector space over the scalar field F is a set V endowedwith two operations, one between vectors: if x, y V then x+y V, andone between scalars and vectors: ifc F andx V thencx Vhaving thefollowing properties:- commutativity of addition: x + y= y + x- associativity of addition: x + (y+ z) = (x + y) + z

- existence of zero: there is an element0 V so thatx + 0 =x for allx V- existence of the opposite: for any x V there is an opposite, denoted byx, so thatx + (x) = 0- distributivity of scalar multiplication with respect to vector addition: c(x +y) =cx + cy- distributivity of scalar multiplication with respect to field addition: (c+d)x= cx + dx- compatibility of scalar multiplication with field multiplication: c(dx) =(cd)x- identity element of scalar multiplication: 1x= x.

Remark 2. From the axioms above we can deduce other properties, which

are obvious for Rn

and Cn

, but we need to prove them in general. Forexample, (i) the zero scalar multiplied by any vector is the zero vector, and(ii) the scalar1 multiplying any vector equals the opposite of the vector.

The proofs of the two facts above are included here for completeness.To show (i) note that 0u = (0 + 0)u = 0u+ 0u so 0u = 0u+ 0u andadding the opposite of 0u we gat 0u = 0. Then to show (ii) note that(1)u + u= (1) + 1u= (1 + 1)u= 0u= 0.


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4 RODICA D. COSTIN

Remark. Another name used for vector spaceis linear space. The lattername is preferable when the space V consists of functions, see examples

4.-8. below.Vector spaces over the scalars F = R are also called vector spaces overthe reals, or real vector spaces, and similar simplified expressions can beused for the complex case F = C

1.2.3. Examples.1. R, R2, R3, . . . Rn are vector spaces over the reals.2. C, C2, C3, . . . Cn are vector spaces over the complex numbers.3. RZ+ ={(x1, x2, x3, . . . , xn, . . .) ; xj R}is a vector space over the reals.4. The set of all polynomials with real coefficients, of degree at most n is avector space over R.

5. The set of polynomials with real coefficients is a vector space overR

.6. The set of polynomials with complex coefficients is a vector space overC.7. The set of continuous functions f : [0, 1] Ris a vector space over R.8. The set of all solutions of the linear differential equation u(t) =u(t) isa vector space.Exercise. Justify the statements 3.-8.

1.3. Subspaces. Let Vbe a vector space over the scalars F.

Definition 3. A subsetUV is called a subspace ofVif sums of elementsof U belong to Uand any scalar multiples of elements of U belong to U.

(More compactly, ifx, y Uthen also x + y U andcx U for allc F.)

Remark. Note that a subspaceUis a vector space. Indeed, addition andmultiplication with scalars of elements in U remain in U, the zero vectorbelongs to Usince ifu Uthen 0u= 0 U, the opposite ofu U belongsto U since u= (1)u U (see Remark 2).

1.3.1. Examples.a. U R2 given by U={(x1, x2) |x1 = 3x2} is a subspace ofR

2.b. U R3, U={(x1, x2, x3) |x1= 3x2, x3= x2} is a subspace ofR

3.c. U R3, U={(x1, x2, x3) |x2= 1} is not a subspace ofR

3.More examples.

1. Lines in the plane, passing through the origin are subspaces in R2.Indeed, lines in the plane are given by equationsAx1 + Bx2= C, which passthrough the origin only for C= 0. Therefore such a line can be written as

U={(x1, x2) R2 |Ax1+ Bx2= 0}

and it is easy to check that such a line is a subspace.2. Lines in space, passing through the origin are subspaces in R3.3. Planes passing through the origin are subspaces in R3.


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LINEAR ALGEBRA 5

Indeed, a plane in R3 is given be a linear equation: Ax1+ Bx2+ Cx3= D.The plane passes through the origin for D= 0. Consider then a plane

U={(x1, x2, x3) R3 |Ax1+ Bx2+ Cx3 = 0}

It is now easy to show that U is a subspace.4. The space of all polynomials with real coefficients, of degree at most nis a subspace of the space of all polynomials with real coefficients.5. The space all polynomials with real coefficients is a subspace of the spaceof all continuous functions defined on [0, 1].

Exercise. Explain why the above examples are indeed subspaces.Counterexample. The set {(x1, x2) R

2 |xj 0} is not a subspace.(Why?)

Definition 4. LetU, Wtwo subspaces ofV. Then their intersectionUW

consists of all vectorx that belong to bothU andW.Exercise. Show that the intersection of two subspaces is also a subspace.Example. Consider two planes U, Win the space, containing the origin

O. IfU=Wthen their intersection is a line containing O.

Definition 5. LetU, Wtwo subspaces ofV. Then their sumU+W consistsof all vectorsu + w withu U andw W.

Exercise. Show that U+ W is also a subspace.Examples.

1. Consider two lines U, W in the space R3, passing thorough the origin. IfU=W then U+ Wis the plane containing the two lines.

2. Consider a line Uand a plane Win the space R3

, both passing thoroughthe origin. IfU is not contained in W then U+W is the whole space R3.But ifUW then U+ W =W.

Exercise. Explain why the examples above are correct.

1.4. Linear Span. Let Vbe a vector space over the scalars F.

Definition 6. A linear combination of some vectors x1, . . . , xr V is avector of the form

c1x1+ . . . + crxr for somec1, . . . , cr F

Note that any linear combination of vectorsx1, . . . , xr Vbelongs to V.Exercise. Show that if all x1, . . . , xr belong to a subspace U ofV, then

any linear combination of these vectors also belongs to the subspace U.

Definition 7. Let x1, . . . , xr be vectors in V. The set ofall linear combi-nations of these vectors is called the subspace spanned byx1, . . . , xr:

Sp(x1, . . . , xr) ={c1x1+ . . . + crxr | c1, . . . , crF}


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Exercise. Show that S p(x1, . . . , xr) is indeed a subspace.Exercise. Show that Sp(x1, . . . , xr) is the smallest subspace containing

x1, . . . , xr.Note: it is useful to define the linear span of infinite sets, too. While thedefinition given above for the linar span is intuitive, it does not generalizeeasily to infinite sets. Here is an equivalent definition which works for finiteand infinite sets as well: the linear span of a set of vectors S (with SV)is the intersection of all subspaces containing S. (The linear span of S istherefore a vector space.)

1.5. Linear independence.

Definition 8. A finite set of vectors x1, . . . , xr V are called linearlydependentif there is a nontrivial linear relation between them.

In other words, x1, . . . , xr Vare called linearly dependentif there exist

some scalars c1, . . . , cr Fnot all zero so that their linear combination isthe zero vector:c1x1+ . . . + crxr = 0

Note that this means that (at least) one of the vectors x1, . . . , xr belongsto the span of the others. (Why?)

Remark 9. A useful observation: two (nonzero) vectors are linearly depen-dent if and only if they are scalar multiples of each other.

Indeed, let u, v V with cu+dv = 0 with not both c, d zero. Say c= 0then dividing by c we have u+ d

cv = 0 and adding the opposite u = d

cv

henceuis a multiple ofv. Furthermore, note that also d = 0 (otherwise wewould have u= 0) hence also v= c

du.

Definition 10. A set of vectors x1, . . . , xr V are called linearly inde-pendent if they are not linearly dependent. In other words, if for somec1, . . . , cr F we have c1x1 + . . . + crxr = 0 then necessarily all c1 =0, . . . , cr = 0.

And more generally:

Definition 11. An (infinite) set of vectorsSV is called linealy indepen-dent if all its finite subsets are linealy independent.

Remark 12. A linearly independent set cannot contain the zero vector.

Indeed, consider a collection x1 = 0, x2, . . . , xr V. Then they arelinearly dependent, since we have c1x1+ c2x2+ . . . + c

rxr

= 0 forc1= 1, c2 = 0, . . . , cr = 0.

1.6. Basis and Dimension.

Definition 13. A set of vectorsS V is a basis ofV if:(i) it is linearly independent and(ii) its span equalsV.


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LINEAR ALGEBRA 7

Theorem 14. Any vector space has a basis. Moreover, all the basis of Vhave the same number of elements, which is called the dimension ofV.

(The proof will not be discussed here.)Remark: if the dimension ofV is infinite, we can also distinguish between

different magnitudes of infinity...but this has to wait until Math 602.

From here on we will mostly discuss finite-dimensional vector

spaces.

Examples

1. Rn is a vector space over R of dimension n, with a basis consisting ofthe vectors

e1= (1, 0, . . . , 0), e2= (0, 1, 0 . . . , 0), . . . , en = (0, 0, . . .0, 1)

Indeed, any x Rn is written as x = (x1, . . . xn) and can be written as alinear combination of them

x= x1e1+ x2e2+ . . . + xnen

Also, the vectors e1, . . . , en are linearly independent.2. Similarly, Cn = {(z1, . . . zn) | zj C} is a vector space over C of

dimension n, with a basis consisting of the vectors(1, 0, . . . , 0), (0, 1, 0 . . . , 0), . . . , (0, 0, . . .0, 1).

3. The set of polynomials of degree at most n, with coefficients in F(which is R or C)

Pn= {p(t) =a0+ a1t + a2t2 + . . . + ant

n |aj F}

is a vector space over the field of scalars F, of dimension n + 1, and a basisis 1, t , t2, . . . tn.4. The set of all polynomials with coefficients in F (which is R or C)

P={p(t) =a0+ a1t + a2t2 + . . . + ant

n | aj F, n N}

is a vector space over the field of scalars F, has infinite dimension.Exercise. Justify the statements in the examples above.

Let V be a vector space over F and let v1, v2, . . . , vn be a basis. SinceSp(v1, v2, . . . , vn) = V then any x V belongs to Sp(v1, v2, . . . , vn), hencehas the form

(1) x= c1v1+ c2v2+ . . . + cnvn for some scalars c1, c2, . . . , cn

which is called the representation of the vector x in the basis v1, v2, . . . , vn.Moreover, the representation is unique. Indeed, suppose thatx can

be represented as (1), and also as

(2) x= d1v1+ d2v2+ . . . + dnvn for some scalars d1, d2, . . . , dn

Then subtracting (1) and (2) we obtain

c1v1+ c2v2+ . . . + cnvn (d1v1+ d2v2+ . . . + dnvn) = 0


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which (using the properties of the operations in a vector space) can bewritten as

(c1 d1)v1+ (c2 d2)v2+ . . . + (cn dn)vn = 0

and sincev1, v2, . . . , vnare linearly independent, then necessarily (c1d1) =0, . . . , (cn dn) = 0 therefore c1= d1, . . . , cn = dn.

Theorem 15. If a vector spaceVhas finite dimensionn, then any collectionconsisting ofn + 1 vectors is linearly dependent.

Why: Let x1, . . . , xnbe a basis for V. Assume, to arrive at a contradiction,that v1, . . . , vn+1 V are linearly independent. Sincex1, . . . , xn is a basisthen

v1=c1x1+ . . . + cnxn for somec1, . . . , cn F

Sincev1= 0 then at least one scalar cj is not zero, say c1= 0. Then we canwrite x1 in terms ofv1, x2, . . . , xn.

We repeat the argument for v2, and so on. In the end we obtain that vn+1is a linear combination ofv1, . . . , vn which contradicts their linear indepen-dence.

Theorem 16. Every linearly independent set of vectorsv1, . . . , vr in a finitedimensional vector spaceVcan be completed to a basis ofV.

How: IfS p(v1, . . . , vr) =V then v1, . . . , vr form a basis and we are done.Otherwise, there is some vector in V, call it vr+1, that cannot be written

as a linear combination ofv1, . . . , vr. Then this means that v1, . . . , vr, vr+1

is a linearly independent set (convince yourselves!).We then repeat the steps above with v1, . . . , vr, vr+1.The procedure must end, since by Theorem 15 we can have at most n

linearly independent vectors.

1.6.1. More examples.1. Let u= (1, 2,3) R3. What is Sp(u)?Solution: Sp(u) ={cu | c R}is the line through O in the direction ofu.2. Let u= (1, 2,3), v= (2, 4,6) R3. What is Sp(u, v)?Solution: note that v = 2u hence Sp(u, v) = Sp(u, 2u) = Sp(u) = as

above.3. Let u= (1, 2,3), w= (1, 1, 0) R3. What is Sp(u, w)?Solution: (think geometrically) ifu, ware dependent, then they are multi-

ple of each other (by Remark 9), and it is obvious by inspection that it is notcase. Hence Sp(u, w) = the smallest subspace containing two independentvectors = the plane (through O) containing them.

4. Let u = (1, 2,3), v = (2, 4,6), w = (1, 1, 0) R3. What isSp(u , v , w)? Find a basis for this subspace. What is its dimension?

Solution:since v = 2u then Sp(u , v , w) = Sp(u, 2u, w) = Sp(u, w) = theplane containing u, w.


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Basis: clearly u, w are independent and span Sp(u , v , w) so they form abasis. Dimension 2.

What if we are not quite sure ifu, w are independent? Lets check: sup-pose that for some scalars c, d we have cu+ dw = 0. But cu+ dw =c(1, 2,3)+d(1, 1, 0) = (c+d, 2c+d,3c) = (0, 0, 0) hencec+d= 0, 2c+d=0 3c= 0 hence c= d = 0, independent!

5. Show that x= (1, 0, 0), y = (1, 1, 0), z = (1, 1, 1) form a basis for R3.Do they form a basis for C3 (as a complex vector space)?

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Lecture 1 Vector Spaces