restart;
with(LinearAlgebra):
Problem 1
a) Find a basis of the rowspace of A,
b) Find a basis of the columnspace of A
c) find a basis of the nullspace of A.
A := Matrix(5, 7, {(1, 1) = 2, (1, 2) = 1, (1, 3) = -1, (1, 4)
= 1, (1, 5) = 4, (1, 6) = 0, (1, 7) = 9, (2, 1) = 2, (2, 2) =
2, (2, 3) = 2, (2, 4) = 0, (2, 5) = 6, (2, 6) = 1, (2, 7) = 7,
(3, 1) = 1, (3, 2) = 1, (3, 3) = 1, (3, 4) = 1, (3, 5) = 5, (3,
6) = 0, (3, 7) = 6, (4, 1) = -2, (4, 2) = 1, (4, 3) = 7, (4, 4)
= 1, (4, 5) = 8, (4, 6) = 1, (4, 7) = -4, (5, 1) = 2, (5, 2) =
1, (5, 3) = -1, (5, 4) = 0, (5, 5) = 2, (5, 6) = 1, (5, 7) = 6}
);
(1)
R := ReducedRowEchelonForm(A);
(2)
Basis of rowspace(A) is
Row(R,1..4);
(3)
Basis of columnspace is
Column(A, [1,2,4,6]);
(4)
(4)
Column(A,5)=-1*Column(A,1)+4*Column(A,2)+2*Column(A,4);
(5)
Basis of nullspace (by hand)
NullSpace(A);
(6)
Problem 2
Do the following vectors span
? If so, cut the list down to a basis.
(v1,v2,v3,v4,v5):=Vector(3, {(1) = 3, (2) = 1, (3) = 1}),
Vector(3, {(1) = 8, (2) = 3, (3) = 2}), Vector(3, {(1) = 2, (2)
= 1, (3) = 0}), Vector(3, {(1) = 30, (2) = 11, (3) = 8}),
Vector(3, {(1) = 16, (2) = 5, (3) = 5});
(7)
v1,v2,v3,v4,v5;
(8)
A := <v1|v2|v3|v4|v5>;
(9)
(9)
ReducedRowEchelonForm(A);
(10)
A basis is v1, v2, and v5
Problem 3
Do the following vectors span
? If so cut the list down to a basis.
v1 := Vector(3, {(1) = 1, (2) = 2, (3) = -1}); v2 := Vector(3,
{(1) = 2, (2) = 1, (3) = 3}); v3 := Vector(3, {(1) = 1, (2) =
-4, (3) = 9});v4 := Vector(3, {(1) = -5, (2) = 2, (3) = -15});
(11)
A := <v1|v2|v3|v4>;
(12)
ReducedRowEchelonForm(A);
(13)
(13)
The conclusion is that
and are a basis for
Problem 4
has dimension 2, and so is not equal to
The vectors
.
Consider the matrix
C := Matrix(4, 4, {(1, 1) = -93,
4) = -98, (2, 1) = -76, (2, 2) =
(3, 1) = -72, (3, 2) = 7, (3, 3)
-2, (4, 2) = 4, (4, 3) = -9, (4,
(1, 2) = -5, (1, 3) = 12, (1,
2, (2, 3) = -4, (2, 4) = -77,
= -16, (3, 4) = 57, (4, 1) =
4) = 27});
(14)
Determinant(C);
13951
(15)
ReducedRowEchelonForm(C);
(16)
Since is invertible, the columns of form a bais for
Now consider the vectors
v1 := Vector(4, {(1) = 1, (2) = 2, (3) = 1, (4) = 1}); v2 :=
Vector(4, {(1) = -1, (2) = 2, (3) = 3, (4) = 2});
(17)
Show that
and
are independent. Complete the list
to a basis by adding on some of the
columns of .
ReducedRowEchelonForm(<v1|v2>);
(18)
A := <v1|v2|C>;
(19)
ReducedRowEchelonForm(A);
(20)
So, a basis is
'v1' = v1, 'v2'=v2, 'Column(C,1)' =Column(C,1), 'Column(C,4)' =
Column(C,4);
(21)
Now complete
to a basis by adding on some of the standard basis vectors
A := <v1|v2|IdentityMatrix(4)>;
(22)
ReducedRowEchelonForm(A);
(23)
(23)
So, the basis is
Problem 5
Consider the vectors
and
given below
a) Cut down the list
to a basis for
Express the other vectors in
the list as linear combinations of the basis vectors.
b) For each of the vectors
and , determine if the vector is in and, if so, express it as a linear
combination of the basis vectors.
v1 := Vector(4, {(1) = 4, (2) = 21, (3) = 120, (4) = 92}); v2
:= Vector(4, {(1) = 5, (2) = 40, (3) = 220, (4) = 98}); v3 :=
Vector(4, {(1) = 7, (2) = 78, (3) = 420, (4) = 110}); v4 :=
Vector(4, {(1) = 1, (2) = 5, (3) = 29, (4) = 23});
(24)
w1 := Vector(4, {(1) = 16, (2) = 97, (3) = 547, (4) = 351});w2
:= Vector(4, {(1) = 17, (2) = 123, (3) = 682, (4) = 349});
(25)
A := <v1|v2|v3|v4|w1|w2>;
(26)
ReducedRowEchelonForm(A);
(27)
A basis for is
and we have
v3 = -2*v1+3*v2;
(28)
The vector
is in and we have
w1 = 2*v1+v2+3*v4;
(29)
The vector
is not in