4.4:3,4,6,12,13,14 4.5:2,8,11,18
DUE: Friday Oct. 27, Beginning of Class
4.4:3) x = 3 [ 1 ] + 0 [ 5 ] + (-1) [ 4 ] = [-1 ]
[-4 ] [ 2 ] [-7 ] [-5 ]
[ 3 ] [-2 ] [ 0 ] [ 9 ]
4.4:4) x = -4 [-1 ] + 8 [ 3 ] + (-7) [ 4 ] = [ 4+24-28 ] = [ 0 ]
[ 2 ] [-5 ] [-7 ] [-8-40+49 ] [ 1 ]
[ 0 ] [ 2 ] [ 3 ] [ 0+16-21 ] [-5 ]
4.4:6) Solve the system c_1 b_1 + c_2 b_2 = x for (c_1,c_2),
represented by the matrix
[ b_1 b_2 x ] = [ 1 5 4 ] [ 1 5 4 ]
[-2 -6 0 ] ~ [ 0 4 8 ]
Thus c_2 = 8/4 = 2, and c_1 = 4 - 5c_2 = -6,
so [x]_B (the representation of x in the basis B) is [ c_1 ] = [-6 ]
[ c_2 ] [ 2 ]
4.4:12) Find [x]_B USING AN INVERSE matrix. Recall x = (P_B)([x]_B)
where P_B is the matrix having columns the vectors in the ordered basis B.
P_B = [ 4 6 ]
[ 5 7 ]
Thus (P_B)^-1 = 1/(28-30) [ 7 -6 ] = -1/2 [ 7 -6 ] = [-7/2 3 ]
[-5 4 ] [-5 4 ] [ 5/2 -2 ]
Alternatively the general procedure to find an inverse when we don't have a
formula (as we did above for 2x2 matrix) is to adjoin P_B to I (the identity)
and row reduce P_B to I to get the matrix [ I (P_B)^-1 ].
[ 4 6 1 0 ] [ 1 3/2 1/4 0 ]
[ 5 7 0 1 ] ~ [ 1 7/5 0 1/5 ] ~
[ 1 3/2 1/4 0 ] [ 1 0 -7/2 3 ] = [ I P_B^-1 ]
[ 0 -1/10 -1/4 1/5 ] ~ [ 0 1 5/2 -2 ]
Now [x]_B = (P_B^-1) x = [-7/2 3 ] [ 2 ] = [-7 ].
[ 5/2 -2 ] [ 0 ] [ 5 ]
4.4:13) [p]_B = (c_1,c_2,c_3), where
c_1 (1 + t^2) + c_2 (t + t^2) + c_3 (1 + 2t + 2t^2) = p(t) = 1 + 4t + 7t^2
Equating like powers of t gives
c_1 + c_3 = 1
c_2 + 2c_3 = 4
c_1 + c_2 + c_3 = 7
The augmented matrix for above system is
[ 1 0 1 1 ] [ 1 0 1 1 ]
[ 0 1 2 4 ] ~ [ 0 1 2 4 ] ~
[ 1 1 1 7 ] [ 0 1 0 6 ]
[ 1 0 1 1 ] [ 1 0 0 2 ]
[ 0 1 2 4 ] ~ [ 0 1 0 6 ]
[ 0 0 -2 2 ] [ 0 0 1 -1 ]
Thus [p]_B = [ 2 ] , the representation of p relative to the basis B.
[ 6 ]
[-1 ]
4.4:14) Again let [p]_B = (c_1,c_2,c_3), where
c_1 (1 - t^2) + c_2 (t - t^2) + c_3 (2 - 2t + t^2) = p(t) = 3 + t - 6t^2
Equating like powers of t gives
c_1 + 2c_3 = 3
c_2 - 2c_3 = 1
-c_1 - c_2 + c_3 = -6
Note this is the same system as [p]_S = P_B [p]_B, where S is the standard
basis for the space of all polynomials of degree <= 2, that is S = {1,t,t^2}.
The augmented matrix for the system is
[ 1 0 2 3 ] [ 1 0 2 3 ]
[ 0 1 -2 1 ] ~ [ 0 1 -2 1 ] ~
[ -1 -1 1 -6 ] [ 0 -1 3 -3 ]
[ 1 0 2 3 ] [ 1 0 0 7 ]
[ 0 1 -2 1 ] ~ [ 0 1 0 -3 ]
[ 0 0 1 -2 ] [ 0 0 1 -2 ]
Thus [p]_B = [ 7 ] , the representation of p in the basis B.
[-3 ]
[-2 ]
4.5:2)
a) V = {[ 4s ] : s,t in R } = { s [ 4 ] + t [ 0 ] : s,t in R }
[ -3s ] [-3 ] [ 0 ]
[ -t ] [ 0 ] [-1 ]
Thus { [ 4 ] [ 0 ] } spans the subspace, and note the vectors are linearly
[-3 ], [ 0 ]
[ 0 ] [-1 ]
independent, so the set is a basis.
b) The dimension of the subspace is the # of vectors in the basis; that is
dim V = 2.
4.5:8) a) V = { (a,b,c,d) : a-3b+c = 0 } = { (a,b,c,d) : c = -a+3b } =
= { [ a ] : a,b,d in R } = { a[ 1 ] + b[ 0 ] + d [ 0 ] : a,b,d in R }
[ b ] [ 0 ] [ 1 ] [ 0 ]
[-a+3b] [-1 ] [ 3 ] [ 0 ]
[ d ] [ 0 ] [ 0 ] [ 1 ]
The three vectors are clearly linearly independent (the only way to get the
zero vector is if a = b = d = 0) and span V.
Thus a basis is { (1,0,-1,0) , (0,1,3,0) , (0,0,0,1) }.
Alternatively, we could write
V = { [3b-c ] : b,c,d in R } = { b[ 3 ] + c[-1 ] + d [ 0 ] : b,c,d in R }
[ b ] [ 1 ] [ 0 ] [ 0 ]
[ c ] [ 0 ] [ 1 ] [ 0 ]
[ d ] [ 0 ] [ 0 ] [ 1 ]
so an alternative basis for V is { (3,1,0,0), (-1,0,1,0), (0,0,0,1) }.
b) dim V = 3.
4.5:11) First note that the 4 vectors are all in R^3, so the dimension of
span { v1, v2, v3, v4 } must be less than or equal to 3.
S = { [ 1 ] , [ 3 ] , [ 9 ] , [-7 ] } = { v1, v2, v3, v4 }
[ 0 ] [ 1 ] [ 4 ] [-3 ]
[ 2 ] [ 1 ] [-2 ] [ 1 ]
By the Spanning Set Thm., some subset of S is a basis for span S. We determine
a subset that is linearly independent. Let
A = [ v1 v2 v3 v4 ] = [ 1 3 9 -7 ]
[ 0 1 4 -3 ] ~
[ 2 1 -2 1 ]
[ 1 3 9 -7 ] [ 1 3 9 -7 ]
[ 0 1 4 -3 ] ~ [ 0 1 4 -3 ]
[ 0 -5 -20 15 ] [ 0 0 0 0 ]
Col A is the same subspace as span S. Reduction to the echelon form above
shows that Columns 1 and 2 are pivot columns.
Hence dim(span S) = dim(Col A) = # of pivot columns of A = 2.
Note { v1, v2 } = { [ 1 ] , [ 3 ] } the set of (original) pivot columns of A,
[ 0 ] [ 1 ] is a basis for span S.
[ 2 ] [ 1 ]
4.5:18) We can see A = [ 1 4 -1 ] is already in an echelon form. A
[ 0 7 0 ]
[ 0 0 0 ]
has 2 pivot columns; also the # free variables for Ax = 0 equals the number of
nonpivot columns which is 1.
Thus dim(Col A) = 2
dim(Nul A) = 1.