Subspace and Dimension: C = {(0,0,0,0), (0,0,1,1), (0,1,0,1), (0,1,1,0)}

[Jenkins99]

Basis and Dimension

Picture:

Given [the equation after "gegeben ist"], does C form a subspace? If so,
what is a basis of C and the dimension?

I determined if the vectors were dependent:

a(0, 0, 0, 0) + b(0, 0, 1, 1) + c(0, 1, 0, 1) + d(0, 1, 1, 0) = (0, 0, 0, 0).

I did away with a(0, 0, 0, 0), which left:

(0, 0, 1, 1) + (0, 1, 0, 1) + (0, 1, 1, 0) = (0, 0, 0, 0).

Column picture:

0 0 0
0 1 1
1 0 1
1 1 0

Which reduces to:

0 1 1
1 0 1
1 1 0

Solving for linear dependence:

c + d = 0
b + d = 0
b + c = 0

c = -d
b = -d
(-d) + (-d) = 0

-2d = 0; trivial solution.

Three linearly independent vectors forming a basis in three dimensions:

(0, 0, 1, 1), (0, 1, 0, 1), (0, 1, 1, 0)

. . .

The answer given in the book is:

C is a subspace since C is closed with regards to addition and
multiplication with a scalar. The dimension of C is 2; a basis is, for
example, (0, 0, 1, 1) and (0, 1, 0, 1).

 

[Jenkins99]

[If this is a duplicate, please delete; I didn't see a confirmation when I submitted several hours ago.]

From the book:

---

8. Gegeben ist

. . . . .\(\displaystyle C\, =\, \left\{\, (0,\, 0,\, 0,\, 0),\, (0,\, 0,\, 1,\, 1),\, (0,\, 1,\, 0,\, 1),\, (0,\, 1,\, 1\, 0)\,\right\}\, \subseteq\, \mathbb{Z}_2^4\)

Bildet C einen Teilraum? Wenn ja, geben Sie eine Basis und die Dimension von C an.

Given [the equation after "gegeben ist"], does C form a subspace? If so, what is a basis of C and the dimension?

---

I determined if the vectors were dependent:

a(0, 0, 0, 0) + b(0, 0, 1, 1) + c(0, 1, 0, 1) + d(0, 1, 1, 0) = (0, 0, 0, 0).

I did away with (0, 0, 0, 0), which left:

(0, 0, 1, 1) + (0, 1, 0, 1) + (0, 1, 1, 0) = (0, 0, 0, 0).

Column picture:

0 0 0
0 1 1
1 0 1
1 1 0

Which reduces to:

0 1 1
1 0 1
1 1 0

Solving for linear dependence:

c + d = 0
b + d = 0
b + c = 0

c = -d
b = -d
(-d) + (-d) = 0

-2d = 0; trivial solution.

Three linearly independent vectors forming a basis in three dimensions:

(0, 0, 1, 1), (0, 1, 0, 1), (0, 1, 1, 0)

. . .

The answer given in the book is:

C is a subspace since C is closed with regards to addition and
multiplication with a scalar. The dimension of C is 2; a basis is, for
example, (0, 0, 1, 1) and (0, 1, 0, 1).

 

[Steven G]

Picture:

Given [the equation after "gegeben ist"], does C form a subspace? If so,
what is a basis of C and the dimension?

I determined if the vectors were dependent:

a(0, 0, 0, 0) + b(0, 0, 1, 1) + c(0, 1, 0, 1) + d(0, 1, 1, 0) = (0, 0, 0, 0).

I did away with a(0, 0, 0, 0), which left:

(0, 0, 1, 1) + (0, 1, 0, 1) + (0, 1, 1, 0) = (0, 0, 0, 0).

Column picture:

0 0 0
0 1 1
1 0 1
1 1 0

Which reduces to:

0 1 1
1 0 1
1 1 0

Solving for linear dependence:

c + d = 0
b + d = 0
b + c = 0

c = -d
b = -d
(-d) + (-d) = 0

-2d = 0; trivial solution.

Three linearly independent vectors forming a basis in three dimensions:

(0, 0, 1, 1), (0, 1, 0, 1), (0, 1, 1, 0)

. . .

The answer given in the book is:

C is a subspace since C is closed with regards to addition and
multiplication with a scalar. The dimension of C is 2; a basis is, for
example, (0, 0, 1, 1) and (0, 1, 0, 1).

d is a numerical value, not a group of words (like trivial solution). Id d=0 the n what does b and c equal??

 

[Ishuda]

Picture:

Given [the equation after "gegeben ist"], does C form a subspace? If so,
what is a basis of C and the dimension?

I determined if the vectors were dependent:

a(0, 0, 0, 0) + b(0, 0, 1, 1) + c(0, 1, 0, 1) + d(0, 1, 1, 0) = (0, 0, 0, 0).

I did away with a(0, 0, 0, 0), which left:

(0, 0, 1, 1) + (0, 1, 0, 1) + (0, 1, 1, 0) = (0, 0, 0, 0).

Column picture:

0 0 0
0 1 1
1 0 1
1 1 0

Which reduces to:

0 1 1
1 0 1
1 1 0

Solving for linear dependence:

c + d = 0
b + d = 0
b + c = 0

c = -d
b = -d
(-d) + (-d) = 0

-2d = 0; trivial solution.

Three linearly independent vectors forming a basis in three dimensions:

(0, 0, 1, 1), (0, 1, 0, 1), (0, 1, 1, 0)

. . .

The answer given in the book is:

C is a subspace since C is closed with regards to addition and
multiplication with a scalar. The dimension of C is 2; a basis is, for
example, (0, 0, 1, 1) and (0, 1, 0, 1).

I believe your book is wrong and you are correct: Take three cases of
a (0,0,0,0) + b (0,0,1,1) + c (0,1,0,1) + d (0,1,1,0)
b = -1/2, c= 1/2, d= 1/2 we have the vector (0,1,0,0)
b = 1/2, c=-1/2, d= 1/2 we have the vector (0,0,1,0)
b = 1/2, c= 1/2, d=-1/2 we have the vector (0,0,0,1)
Obviously a basis set of {(0,1,0,0), (0,0,1,0), (0,0,0,1)} forms a three space.