4.1:4,10,14,21 4.2:4,11,24,31 4.3:5,10,24
DUE: Friday Oct. 20, Beginning of Class
4.1:4) Take any two (non-identical) vectors u,v on any line not through the
origin (graph any such line). They determine a parallelogram. Their sum vector
u + v is the diagonal vector of the parallelogram, not on the line so not
closed under vector addition.
4.1:10)
H = { [ 2t ] , t in R } = span { [ 2 ] }
[ 0 ] [ 0 ]
[ -t ] [-1 ]
is a subspace of R^3 by Theorem 1.
4.1.14) w is in span {v_1,v_2,v_3} iff w can be written as a linear combination
of v_1,v_2,v_3. Equivalently w is in span {v_1,v_2,v_3} iff the system Ax = w,
where A = [ v_1 v_2 v_3 ], has a solution (is consistent). The augmented
matrix [ A w ] is
[ 1 2 4 8 ] [ 1 2 4 8 ]
[ 0 1 2 4 ] ~ [ 0 1 2 4 ] ~
[ -1 3 6 7 ] [ 0 5 10 15 ]
[ 1 2 4 8 ] [ 1 2 4 8 ]
[ 0 1 2 4 ] ~ [ 0 1 2 4 ]
[ 0 1 2 3 ] [ 0 0 0 -1 ]
The last row says "0 = -1", a contradiction, so the system is inconsistent and
thus w is not in span {v_1,v_2,v_3}.
4.1:21) Yes, H = { [ a b ] : a,b,d in R }
[ 0 d ]
is a subspace of M_2x2 (the vector space of all 2x2 matrices). Using the
definition of subspace, you need to show three properties:
a) Take a = b = d = 0. Then
[ 0 0 ] , the zero element of M_2x2, is in H.
[ 0 0 ]
b) Let A_1 = [ a_1 b_1 ] and A_2 = [ a_2 b_2 ]
[ 0 d_1 ] [ 0 d_2 ]
Then A_1 + A_2 = [ (a_1+a_2) (b_1+b_2) ]
[ 0 (d_1+d_2) ]
which is in H since of same form.
c) For any c in R, c [ a b ] = [ ca cb ] is in H.
[ 0 d ] [ 0 cd ]
Alternative solution:
H = span{ [ 1 0 ] , [ 0 1 ] , [ 0 0 ] }
[ 0 0 ] [ 0 0 ] [ 0 1 ]
so by Thm. 1, H is a subspace of M_2x2.
4.2:4) The augmented matrix of Ax = 0 (we find all solutions of this
homogeneous system to determine Nul(A)) is
[ 1 -6 4 0 0 ] [ 1 -6 0 0 0 ]
[ 0 0 2 0 0 ] ~ [ 0 0 1 0 0 ]
Thus x_2, x_4 free
x_3 = 0
x_1 = 6x_2
Hence [ x_1 ] = [ 6x_2 ] = x_2 [ 6 ] + x_4 [ 0 ]
[ x_2 ] [ x_2 ] [ 1 ] [ 0 ]
[ x_3 ] [ 0 ] [ 0 ] [ 0 ]
[ x_4 ] [ x_4 ] [ 0 ] [ 1 ]
Thus Nul(A) = span { (6,1,0,0), (0,0,0,1) }. Note that the # of vectors in the
spanning set equals the # of free variables for the system Ax = 0.
4.2:11) W = { [ 0 ] + b [ 1 ] + d [-2 ] : b,d real }
[ 5 ] [ 0 ] [ 1 ]
[ 0 ] [ 1 ] [ 3 ]
[ 0 ] [ 0 ] [ 1 ]
W is not a vector space since it is not a subspace of R^4 since the zero
vector of R^4 [the vector (0,0,0,0)] is not in W. To see this, we show that
[ 0 ] + b [ 1 ] + d [-2 ] = [ 0 ]
[ 5 ] [ 0 ] [ 1 ] [ 0 ]
[ 0 ] [ 1 ] [ 3 ] [ 0 ]
[ 0 ] [ 0 ] [ 1 ] [ 0 ]
does not have a solution for any b and d in R. This is clear since row4
implies d = 0; which then contradicts row2 (" 5 = 0 ").
Alternatively we can show by row operations that the augmented matrix
[ 1 -2 0 ] [ 1 -2 0 ] [ 1 0 0 ]
[ 0 1 -5 ] [ 0 1 -5 ] [ 0 0 -5 ]
[ 1 3 0 ] ~ [ 0 1 0 ] ~ [ 0 1 0 ]
[ 0 1 0 ] [ 0 1 0 ] [ 0 0 0 ]
is inconsistent, since row2 says "0 = -5', a contradiction.
4.2:24) w is in Col(A) iff Ax = w is consistent. Row reduce augmented matrix:
[ -8 -2 -9 2 ] [ 0 -2 -1 -2 ]
[ 6 4 8 1 ] ~ [ 0 4 2 4 ] ~
[ 4 0 4 -2 ] [ 1 0 1 -1/2 ]
[ 1 0 1 -1/2 ] [ 1 0 1 -1/2 ]
[ 0 4 2 4 ] ~ [ 0 1 1/2 1 ]
[ 0 -2 -1 -2 ] [ 0 0 0 0 ]
Consistent; so w is in Col(A).
w is in Nul(A) iff Aw = 0. Easiest to check if Aw = 0 is true:
Aw = [ -8 -2 -9 ] [ 2 ] = [ -16-2+18 ] = [ 0 ]
[ 6 4 8 ] [ 1 ] [ 12+4-16 ] [ 0 ]
[ 4 0 4 ] [-2 ] [ 8+0-8 ] [ 0 ]
Thus w is in Nul(A).
Alternatively, by row operations on Ax = 0, we get Nul(A) = { x: Ax = 0 } =
{ x_3 [ -1 ] : x_3 real }. Let x_3 = -2, we see -2 [ -1 ] = [ 2 ] = w, so w is
[-1/2] [-1/2] [ 1 ] in Nul(A)
[ 1 ] [ 1 ] [-2 ]
4.2:31)
a) Let p,q be arbitrary polynomials in P_2 (the set of all polynomials of
degree <= 2 )
p = a_0 + a_1 t + a_2 t^2
q = b_0 + b_1 t + b_2 t^2
Note p + q = (a_0 + b_0) + (a_1 + b_1)t + (a_2 + b_2)t^2
Then T(p+q) = [ (p+q)(0) ] = [ a_0 + b_0 ]
[ (p+q)(1) ] [ a_0 + b_0 + a_1 + b_1 + a_2 + b_2 ]
and T(p) + T(q) = [ a_0 ] + [ b_0 ] = T(p+q)
[ a_0+a_1+a_2 ] [ b_0+b_1+b_2 ]
Also note for any real number c
T(cp) = [ cp(0) ] = [ ca_0 ] = c [ a_0 ] = cT(p)
[ cp(1) ] [ ca_0+ca_1+ca_2 ] [ a_0+a_1+a_2 ]
Therefore T is a linear transformation.
b) kernel(T) = { p in P_2 : T(p) = [ 0 ] }
[ 0 ]
T(p) = [ 0 ] iff [ 0 ] = [ p(0) ] = [ a_0 ]
[ 0 ] [ 0 ] [ p(1) ] [ a_0+a_1+a_2 ]
Hence we require a_0 = 0, and
a_1 + a_2 = 0 --> a_1 = -a_2
Thus kernel(T) is { a_0 + a_1 t + a_2 t^2 : a_0 = 0, a_1 = -a_2 }
= { a_1 t - a_1 t^2 : a_1 real } = span{ t - t^2 }
Thus a desired polynomial that spans kernel(T) is p(t) = t - t^2.
range(T) = { T(p) : p in P_2 } = { [ a_0 ] : a_0,a_1,a_2 real } = R^2
[a_0+a_1+a_2 ]
If this is not immediately obvious to you,
to see this, let [ x ] be any vector in R^2 and show that it is in range(T).
[ y ]
We can in particular let a_0 = x, a_1 = y - x, a_2 = 0 to get
[ a_0 ] = [ x ] = [ x ]
[ a_0+a_1+a_2 ] [ x + (y-x) + 0 ] [ y ]
Thus range(T) = R^2, i.e. T is an onto transformation.
4.3:5) Not a basis for R^3 since there are 4 vectors (4 > 3); also since [ 0 ]
is zero vector vector in set impies set is linearly dependent. [ 0 ]
[ 0 ]
Now row reduce the matrix [ v_1 v_2 v_3 v_4 ] =
[ 1 -2 0 0 ] [ 1 -2 0 0 ] [ 1 0 0 0 ]
[-3 9 0 -3 ] ~ [ 0 3 0 -3 ] ~ [ 0 1 0 0 ]
[ 0 0 0 5 ] [ 0 0 0 1 ] [ 0 0 0 1 ]
Note linearly independent columns vectors of the original set are those which
correspond to the three pivot columns. That is, v_1, v_2, v_4 are linearly
independent vectors in R^3, and also span R^3.
4.3:10) Find a basis for the null space of given matrix. Solve Ax = 0
[ 1 0 -5 1 4 0 ] [ 1 0 -5 1 4 0 ] [ 1 0 -5 0 7 0 ]
[-2 1 6 -2 -2 0 ] ~ [ 0 1 -4 0 6 0 ] ~ [ 0 1 -4 0 6 0 ]
[ 0 2 -8 1 9 0 ] [ 0 0 0 1 -3 0 ] [ 0 0 0 1 -3 0 ]
Thus x_5, x_3 free since columns 3 and 5 are not pivot columns.
[ x_1 ] = [ 5x_3 - 7x_5 ] = x_3 [ 5 ] + x_5 [-7 ]
[ x_2 ] [ 4x_3 - 6x_5 ] [ 4 ] [-6 ]
[ x_3 ] [ x_3 ] [ 1 ] [ 0 ]
[ x_4 ] [ 3x_5 ] [ 0 ] [ 3 ]
[ x_5 ] [ x_5 ] [ 0 ] [ 1 ]
>From previous section we know these vectors are linearly independent (without
having to check for independence). Thus
{ (5,4,1,0,0), (-7,-6,0,3,1) } is a basis for Nul(A).
4.3:24)
B = { v_1,v_2,...,v_n } is given to be a linearly independent set of n
vectors in R^n. Define the matrix
A = [ v_1 v_2 ... v_n ] [ 1 0 . . . 0 ]
[ 0 1 0 . . 0 ]
[ 0 0 1 . . . ]
~ [ . . . . . . ] ~ I in reduced
[ . . . . . 0 ] echelon form
[ 0 0 . . . 1 ]
There are n pivot positions (every row) so columns of A span R^n.
Alternatively, by Invertible Matrix Theorem (note A is square), the columns
linearly independent imply that the columns span R^n. Thus B satisfies
definition of basis (linearly indep. set that spans R^n).