Answer :
Let's determine the dimension of the subspaces [tex]\( W \)[/tex] for each given condition in [tex]\( R^4 \)[/tex].
### Exercise 15
Condition: [tex]\( x_1 - 2 x_2 + x_3 - x_4 = 0 \)[/tex]
This equation is a single linear constraint on the four variables [tex]\( x_1, x_2, x_3, x_4\)[/tex]. Thus, the system has [tex]\( 4 - 1 = 3 \)[/tex] free variables.
Dimension of [tex]\( W \)[/tex]: [tex]\( \operatorname{dim}(W) = 3 \)[/tex]
### Exercise 16
Condition: [tex]\( x_1 - 2 x_3 = 0 \)[/tex]
This can be rewritten as [tex]\( x_1 = 2x_3 \)[/tex]. This is a single linear constraint on the four variables. Thus, the system has [tex]\( 4 - 1 = 3 \)[/tex] free variables.
Dimension of [tex]\( W \)[/tex]: [tex]\( \operatorname{dim}(W) = 3 \)[/tex]
### Exercise 17
Condition: [tex]\( x_1 = -x_2 + 2 x_4 \)[/tex]
This condition expresses [tex]\( x_1 \)[/tex] in terms of [tex]\( x_2 \)[/tex] and [tex]\( x_4 \)[/tex]. Thus, it is a single constraint, leaving us with [tex]\( 4 - 1 = 3 \)[/tex] free variables.
Dimension of [tex]\( W \)[/tex]: [tex]\( \operatorname{dim}(W) = 3 \)[/tex]
### Exercise 18
Conditions:
[tex]\[ \begin{cases} x_1 + x_3 - 2 x_4 = 0 \\ x_3 = -x_4 \\ x_2 + 2 x_3 - 3 x_4 = 0 \end{cases} \][/tex]
Let's simplify these conditions:
1. Substitute [tex]\( x_3 = -x_4 \)[/tex] into [tex]\( x_1 + x_3 - 2 x_4 = 0 \)[/tex] and [tex]\( x_2 + 2 x_3 - 3 x_4 = 0 \)[/tex]:
- [tex]\( x_1 - x_4 - 2 x_4 = 0 \)[/tex] ⟹ [tex]\( x_1 = 3 x_4 \)[/tex]
- [tex]\( x_2 - 2 x_4 - 3 x_4 = 0 \)[/tex] ⟹ [tex]\( x_2 = 5 x_4 \)[/tex]
The equations [tex]\( x_1 = 3 x_4 \)[/tex], [tex]\( x_2 = 5 x_4 \)[/tex], and [tex]\( x_3 = -x_4 \)[/tex] express [tex]\( x_1, x_2, \)[/tex] and [tex]\( x_3 \)[/tex] in terms of [tex]\( x_4 \)[/tex]. So, there is only one free variable, [tex]\( x_4 \)[/tex].
Dimension of [tex]\( W \)[/tex]: [tex]\( \operatorname{dim}(W) = 1 \)[/tex]
### Exercise 19
Condition: [tex]\( x_1 = -x_4 \)[/tex]
This condition is a single linear constraint on the four variables [tex]\( x_1, x_2, x_3, x_4\)[/tex]. Thus, the system has [tex]\( 4 - 1 = 3 \)[/tex] free variables.
Dimension of [tex]\( W \)[/tex]: [tex]\( \operatorname{dim}(W) = 3 \)[/tex]
### Exercise 20
Conditions:
[tex]\[ \begin{cases} x_1 - x_2 = 0 \\ x_2 = 3 x_4 \\ x_2 - 2 x_3 = 0 \\ x_3 = 2 x_4 \\ x_3 - x_4 = 0 \end{cases} \][/tex]
Let's simplify these:
1. [tex]\( x_1 = x_2 \)[/tex]
2. [tex]\( x_2 = 3 x_4 \)[/tex]
3. [tex]\( x_2 = 2 x_3 \)[/tex]
4. [tex]\( x_3 = 2 x_4 \)[/tex]
5. [tex]\( x_3 = x_4 \)[/tex]
From the equations above:
- [tex]\( x_2 = 3 x_4 \)[/tex]
- [tex]\( x_3 = x_4 \)[/tex]
The equations turn out to be inconsistent since [tex]\( 2 x_4 = x_4 \)[/tex] only holds if [tex]\( x_4 = 0 \)[/tex]. Thus, [tex]\( x_2 = 0 \)[/tex] and [tex]\( x_3 = 0 \)[/tex] as well. Therefore, [tex]\( x_1 = 0 \)[/tex], so the only solution is the zero vector.
Dimension of [tex]\( W \)[/tex]: [tex]\( \operatorname{dim}(W) = 0 \)[/tex]
### Exercise 15
Condition: [tex]\( x_1 - 2 x_2 + x_3 - x_4 = 0 \)[/tex]
This equation is a single linear constraint on the four variables [tex]\( x_1, x_2, x_3, x_4\)[/tex]. Thus, the system has [tex]\( 4 - 1 = 3 \)[/tex] free variables.
Dimension of [tex]\( W \)[/tex]: [tex]\( \operatorname{dim}(W) = 3 \)[/tex]
### Exercise 16
Condition: [tex]\( x_1 - 2 x_3 = 0 \)[/tex]
This can be rewritten as [tex]\( x_1 = 2x_3 \)[/tex]. This is a single linear constraint on the four variables. Thus, the system has [tex]\( 4 - 1 = 3 \)[/tex] free variables.
Dimension of [tex]\( W \)[/tex]: [tex]\( \operatorname{dim}(W) = 3 \)[/tex]
### Exercise 17
Condition: [tex]\( x_1 = -x_2 + 2 x_4 \)[/tex]
This condition expresses [tex]\( x_1 \)[/tex] in terms of [tex]\( x_2 \)[/tex] and [tex]\( x_4 \)[/tex]. Thus, it is a single constraint, leaving us with [tex]\( 4 - 1 = 3 \)[/tex] free variables.
Dimension of [tex]\( W \)[/tex]: [tex]\( \operatorname{dim}(W) = 3 \)[/tex]
### Exercise 18
Conditions:
[tex]\[ \begin{cases} x_1 + x_3 - 2 x_4 = 0 \\ x_3 = -x_4 \\ x_2 + 2 x_3 - 3 x_4 = 0 \end{cases} \][/tex]
Let's simplify these conditions:
1. Substitute [tex]\( x_3 = -x_4 \)[/tex] into [tex]\( x_1 + x_3 - 2 x_4 = 0 \)[/tex] and [tex]\( x_2 + 2 x_3 - 3 x_4 = 0 \)[/tex]:
- [tex]\( x_1 - x_4 - 2 x_4 = 0 \)[/tex] ⟹ [tex]\( x_1 = 3 x_4 \)[/tex]
- [tex]\( x_2 - 2 x_4 - 3 x_4 = 0 \)[/tex] ⟹ [tex]\( x_2 = 5 x_4 \)[/tex]
The equations [tex]\( x_1 = 3 x_4 \)[/tex], [tex]\( x_2 = 5 x_4 \)[/tex], and [tex]\( x_3 = -x_4 \)[/tex] express [tex]\( x_1, x_2, \)[/tex] and [tex]\( x_3 \)[/tex] in terms of [tex]\( x_4 \)[/tex]. So, there is only one free variable, [tex]\( x_4 \)[/tex].
Dimension of [tex]\( W \)[/tex]: [tex]\( \operatorname{dim}(W) = 1 \)[/tex]
### Exercise 19
Condition: [tex]\( x_1 = -x_4 \)[/tex]
This condition is a single linear constraint on the four variables [tex]\( x_1, x_2, x_3, x_4\)[/tex]. Thus, the system has [tex]\( 4 - 1 = 3 \)[/tex] free variables.
Dimension of [tex]\( W \)[/tex]: [tex]\( \operatorname{dim}(W) = 3 \)[/tex]
### Exercise 20
Conditions:
[tex]\[ \begin{cases} x_1 - x_2 = 0 \\ x_2 = 3 x_4 \\ x_2 - 2 x_3 = 0 \\ x_3 = 2 x_4 \\ x_3 - x_4 = 0 \end{cases} \][/tex]
Let's simplify these:
1. [tex]\( x_1 = x_2 \)[/tex]
2. [tex]\( x_2 = 3 x_4 \)[/tex]
3. [tex]\( x_2 = 2 x_3 \)[/tex]
4. [tex]\( x_3 = 2 x_4 \)[/tex]
5. [tex]\( x_3 = x_4 \)[/tex]
From the equations above:
- [tex]\( x_2 = 3 x_4 \)[/tex]
- [tex]\( x_3 = x_4 \)[/tex]
The equations turn out to be inconsistent since [tex]\( 2 x_4 = x_4 \)[/tex] only holds if [tex]\( x_4 = 0 \)[/tex]. Thus, [tex]\( x_2 = 0 \)[/tex] and [tex]\( x_3 = 0 \)[/tex] as well. Therefore, [tex]\( x_1 = 0 \)[/tex], so the only solution is the zero vector.
Dimension of [tex]\( W \)[/tex]: [tex]\( \operatorname{dim}(W) = 0 \)[/tex]
