Merced College; Don Power

 

LINEAR ALGEBRA - CHAPTER 5, LECTURE

 

5.1  Real Vector Spaces              Hwk:  1, 5, 9, 11, 14, 18, 22, 25, 29, 31, 32

 

Def of a vector space:  set V of objects on which two operations are defined, "addition" and "scalar multiplication," such that the following 10 properties hold:

      addition:  #1 closed, #3 associative, #4 identity, #5 inverse, #2 commutative

      scalar mult:  #6 closed, #9 assoc (mult by 2 scalars ), #10 identity (of scalar element 1)

      distributive (#7 scalar over sum of vectors, #8 vector over sum of scalars)

 

Notice:  These are really the same properties as the properties of vectors in section 4.1, properties of vectors in Euclidean n-space, with the addition of the closure axioms.

 

Examples

      Rn   Ordinary vectors in n-space

            Special cases R, R2, R3, ...

      2´2 matrices with usual matrix addition and scalar mult

      m´n matrices ditto   Mmn

      real-valued functions defined on the entire real number line:  F(-∞,∞)

            Some subsets of these that are functions:

                  real-valued functions defined on an open or closed interval F(a,b) or F[a,b]

                  polynomials of degree £ n  Pn

      set of planes through the origin

      the zero vector space

Non-example:  Ex 5 in text, where ku = (ku1,0)  Axiom 10 fails:  1(u1,u2) = (u1,0), not (u1,u1) as required.

For class:  exercises 3,9:  Which axioms fail and why?

 

5.2  Subspaces
Hwk:  1, 2b, 3b, 4ce, 5ab, 6, 7b, 8b, 9b, 10a, 11b, 12ac, 13, 14c, 15, 16, 20, 23, 24a, 27

 

Subspaces – def:  a subset which is, itself, a vector space.

Thm 5.2.1:  Let W be a [nonempty]subset of V.  W is a subspace iff W is closed under addition and scalar mult. (i.e. u+v is in W and ku is in W whenever u and v are in W and k is any scalar)

Examples and non-examples of subspaces

            (not)  points in a plane that is not through (0,0,0):  Ex x+2y-3z+1=0

            (yes)  polynomials of degree £ n (Ex P2 is subspace of P4)

            (no) [Ex3]  {(x,y)|x³0, y³0}  + is OK but scalar multiplication is not [if k is a negative number].

(yes)  functions with continuous nth derivatives, as subspace of functions continuous on R, as a subspace of real-valued functions on R:   In fact, Pn < C¥ < Cn(-¥,¥) < C(¥,¥) < F(-¥,¥)

Thm 5.2.2  For a homogeneous system of linear equations in n unknowns, the set of solution vectors is a subspace of Rn

      Proof in text.  Related ideas:  For nxn, nontrivial solutions are parameterized®  no unique sol                       ®det(A)=0®image of TA is not all of Rn.

Lin Comb.  Def: w is a linear combination of vectors v1, v2, ..., vr if there exist scalars k1, k2, ..., kr such that w = k1v1 + k2v2 + ... + krvr.

 

      Ex 8;  vectors in R3 are linear comb of "std basis vectors" i,j,k

      Ex 9 Show that a given w is or is not a linear comb of given vectors u & v

 

Span:  [for vectors in a vector space V] Def:  span(S) = the set of all linear combs of a set of vectors S = {v1,v2,...,vr}

 

Thm 5.2.3  W=span(S) is a subspace of V;  W is the smallest subspace of V containing S (any other subspace containing S must contain W)

 

Ex 10:  Space spanned by a single vector (with initial point at the origin) in R2 or R3 is a line

            Space spanned by a pair of non-parallel vectors in R3 (with initial points at the origin) is a plane

 

Ex 11:  Span Pn = {1, x, x2, ... , xn} since all polynomials are linear combs of these vectors

 

Ex 12:  Determining whether a set of vectors span a space:  Determine whether an arbitrary vector  (b1,b2,b3) is a linear comb of the set for some arbitrary set k1, k2, k3.  Set up system of equations Ak = b; 

      if consistent, the vectors span R3, if inconsistent, they do not span.  Note that the columns of A are the given set of vectors.

 

 

5.3  Linear Independence
Hwk:  1, 2bd, 4bd, 5a, 7a, 7b for v1, 9, 10 for {v1,v2}, 12, 14, 15, 19, 20acd, 21bcd, 24

 

Def independent:  only sol. to linear comb = 0 is all-zero

      dependent -- other sols exist (parameterized:  for future, note how many parameters are required)

Ex 1,2  dependent:  Find linear comb that = 0

            Independent set:  X4a

            Dependent set:  use v1=[5,2,3], v2=[-3,1,5], v3=[3,10,29] as example of a dep set,

Ex3            ei's are independent (or {i,j,k})

Ex4            determine whether dependent or independent:  rref yields I -- independent;   or detA=0 -- dependent

Thm 1  Set is Lin Dependent iff at least one of the vectors in S is expressible as a LC of the others;

                  Otherwise, Lin Independent

Ex 6-    Dependent if one is LC of the others.

Thm 2  (a)  A set that contains the zero vector is linear dep.  Pr:  0v1 + 0v2 + ... + 0vr + 10 = 0

            (b)  A set with exactly 2 vectors is Lin Independent iff neither is a scalar mult of the other

                  Demo:  v1 = kv2 implies 1v1 - kv2 = 0

Thm3   Let S = set of r vectors in Rn.  If r>n, S is linear dependent  (ie nr vectors > dimension of the space)

Lin dependence of functions:  Func are linear dependent iff Wronskian = 0 for all x

      Ex:  {1,x-1,x2+x} is Lin Independent

      Ex:  {eix, sin x, cos x} is Lin Dependent

 

 

5.4  Basis and Dimension
Hwk:  1b, 2bd, 3ab, 4c, 5, 6a, 9b, 10c, 11, 13, 16, 17, 18b, 19b, 21b, 24ab, 29, 33a, 34

 

Def:  If V is a vector space and S={v1, v2, ...vn} is a set of vectors in V, S is a basis of V if

      1.  S is linearly independent (LI), and

      2.  S spans V

 

Ex 3 demonstrates that a given set of vectors is a basis:  Showing that A-1 exists is sufficient for both

 

Thm 1:  Given v and a basis S, v can be written as a linear combination (LC) of vectors in S in exactly one way

 

Coordinate vector (v)S:  Write v as a LC of vectors in S;

(v)S is the vector whose elements are the coeffs of v1, v2, ..., vn ("coordinates relative to S").

      Note:  (v)S depends on the choice of vectors in S and the sequence in which they are written.

      Ex 4:  given v, find (v)S and  vice versa.

 

Basis of a subspace:  If S is a LI set of vectors in V, then S is a basis of the subspace span(S)  (Ex 7)

 

Def:  Dimension of a vector space V is the number of vectors in a basis for V

      Thm 2:  (For a finite-dimensional vector space).  If any basis contains n vectors, then:

            Any set with more than n vectors is linearly dependent (at least one must be LC of the others)

            Any set with fewer than n vectors cannot span the vector space

      Thm 3:  all bases contain the same number of vectors

 

Examples of "standard bases" and the dimemsions of their vector spaces:

      i, j, k in R3                                          dim R3 = 3

      e1, e2, ..., en in Rn                      dim Rn = n

      1, x, x2, x3, ..., xn in Pn               dim Pn = n+1

      , , , , ,  in M23    dim Mmn = m´n

Vector spaces may also be infinite-dimensional, e.g. C(-¥,¥), P¥

 

See Ex. 3:  How to demonstrate that a given set of vectors is a basis

      1.   Show lin indep:  Show c1v1 + c2v2 + ... +cnvn = 0 has only the trivial sol.

            How?  Form matrix of col vectors, check rref = I; or take det (not = 0); or verify A inverse exists.

      2.   Show that the set spans V:  Show c1v1 + c2v2 +... +cnvn = col vector with elts b1, b2, ..., bn

            How?  Same way.

      So, checking the coefficient matrix described above verifies both requirements.

      [Notice that this will require that you have a square matrix

 

See Example 10:  Finding a basis of the solution space of a homogeneous system (also called the nullspace): If the solution is trivial, there are no basis vectors, and the dimension of the nullspace is 0.  However if nontrivial solutions exist, they will be parameterized;  ex 10 shows how to write the general solution as a linear combination of basis vectors.  The number of parameters corresponds to the number of basis vectors; that number is the dimension of the nullspace. Steps:

      1.   Write the parameterized solution

      2.   Expand the parameterized solution as a linear combination, where the parameters are the scalar multiples, and the vectors are purely numerical.  These are then the basis vectors.

 

5.5  Row Space, Column Space, and Nullspace
Hwk:  1, 2b, 3b, 4ab, 5b, 6bc, 7b, 8bc, 9bc, 10bc, 11b, 12b, 13, 16b, 18

 

Def:  Nullspace(A) is solution space of the homogeneous syst Ax=0

 

Thm 1:   Ax=b is consistent iff b is in the col space of A

 

Thm 2:  Let x0 be any sol of Ax=b and let {v1, v2, ..., vk} be a basis for the nullspace (i.e. of solution space of Ax=0).  Then

            1.   x0 + any linear combination (of v1, v2, ..., vk) is also a solution of Ax=b, and

            2.   Every solution of Ax=b can be written as x0 + some linear combination (of v1, v2, ..., vk)

 

Thm 3:  Elementary row ops do not change the nullspace of a matrix

 

Thm 4:  Elementary row ops do not change the row space of a matrix

 

What is missing?  Elementary row ops do change the column space of a matrix

 

Thm 5:  If A and B are row equivalent matrices:

1.  A set of column vectors in A are linearly independent iff the corresponding column vectors in B are linearly independent

2.  A set of column vectors in A is a basis for the column space, iff the corresponding column vectors in B are a basis

            3.  The same thing is not true for bases of the row space.

 

Thm 6:  If a matrix R is in row-echelon form, then row vectors with leading 1's are basis for row space

            and column vectors with leading 1's of the row vectors are a basis for the column space.

 

Finding any vectors that form a basis for the row space and column space of A

      Take ref or rref (results will be different, but either way, you still get bases)

            The rows with leading  1's, of ref(A) or of rref(A), are a basis for the row space

      Which columns that contain the same leading 1's?   Those same columns from A are a basis for the column space

             

Finding cols and rows of A that form basis of row space and column space:

      Cols:  take ref(A) or rref(A, identify columns with leading 1s, take these columns from A

      Rows:  take ref(AT), identify columns with leading 1s, take corresponding rows from A

      Why?  Thm 5.5.5b:  If A and B are row equiv, then:

            the same columns of A that form basis for column space of A also form basis for column space of B

 

5.6  Rank and Nullity       Hwk:  1, 2d, 3d, 4bdg, 5, 6, 7bdf, 8bdf, 9, 11, 13, 16, 17ab, 18ab, 19ad

 

Rank(A) = common dimension of row space and col space (Thm 5.6.1:  these dimensions are the same)

      = nr of leading 1's in ref(A) or rref(A)

      = nr of leading variables in solution of Ax=0            Thm 5.6.4(a)

      = rank(AT)                                                             Thm 5.6.2

 

Nullity(A) = dim(null space)            Def

      = nr of parameters of solution of homogeneous system Ax = 0         Thm 5.6.4(b)

      = nr of free variables in ref(A) or rref(A)

 

Thm 5.6.3:  Dimension Thm:  rank(A) + nullity(A) = n         for a matrix with n columns

      So (Thm 5.6.7):  For a consistent system Ax=b, number of parameters (nullity) = n - r  (cols - rank)

 

rank(A) £ min(m,n):  because dim(row sp) = dim(col sp)

 

Thm 5.6.5:  Consistency Thm: for system Ax=b of m equations in n unknowns (hence A is m´n):

      These are equivalent:

            Ax=b is consistent        (i.e. for some matrix m´1 matrix b)

            b is in the column space of A

            A and the augmented matrix [A|b] have the same rank

Thm 5.6.6:  [Same conditions]:  for system Ax=b of m equations in n unknowns (hence A is m´n):

      These are equivalent:

            Ax=b is consistent for every m´1 matrix b

            Col vectors of A span Rm

            Rank(A) = m

Overdetermined system:  more equations than unknowns (m>n)

      Column vectors cannot span Rm

      Therefore Ax=b cannot be consistent for every possible b

            Ex 5:  solutions are linear functions of 2 parameters

Underdetermined system:  fewer equations than unknowns (m<n)

      General solution has at least one parameter (Thm 7:  Number of parameters = n - r with n > m and m ³ r

      Conclusion:  if consistent, Ax=b has infinitely many solutions.

 

 

Return to:  Merced College; Don Power               Updated 04/02/07 by Don Power