Let V = {(a, b, c, d) ( (4: b – 2c + d = 0}
W = {(a, b, c, d) ( (4: a = d, b = 2c}
Find a basis and dimension of
a) V b) W c) V intersection W.
a) To find a basis for V, we need to solve the equation given in V: b - 2c + d = 0.
We can parametrize the solution by letting c and d be free variables.
Let c = t and d = s, where t and s are real numbers.
Then, from the equation b - 2c + d = 0, we can solve for b:
b = 2c - d = 2t - s.
Therefore, a basis for V is {(a, 2t - s, t, s)} where t and d are real numbers.
The dimension of V is 2, since there are two free variables (t and s) in the parametrization.
b) To find a basis for W, we need to solve the equations given in W: a = d and b = 2c.
From the equation a = d, we can set d = s, where s is a real number.
From the equation b = 2c, we can set c = t, where t is a real number.
Then, we can express a and b in terms of s and t:
a = s and b = 2t.
Therefore, a basis for W is {(s, 2t, t, s)} where t and s are real numbers.
The dimension of W is also 2, since there are two free variables (t and s) in the parametrization.
c) To find the intersection of V and W, we need to find the common solutions to the equations given in V and W.
From the equations in V, we have b - 2c + d = 0 and a = d.
From the equations in W, we have a = d and b = 2c.
We can combine these equations:
a = d,
b = 2c,
b - 2c + d = 0.
From the equations a = d and b = 2c, we can express a and b in terms of c:
a = c and b = 2c.
Substituting these values into the equation b - 2c + d = 0:
2c - 2c + d = 0
d = 0.
Therefore, the only common solution to V and W is (c, 2c, c, 0).
A basis for the intersection of V and W is {(c, 2c, c, 0)} where c is a real number.
The dimension of the intersection of V and W is 1, since there is only one free variable (c) in the parametrization.
To find the basis and dimension of each set, we need to determine the linearly independent vectors in each set and count how many there are.
a) V:
To find the basis of V, we need to find the linearly independent vectors that span the set. First, let's rewrite the equation given as a linear combination of vectors:
b - 2c + d = 0
This equation can be rewritten as:
b = 2c - d
Now, we can express V as a linear combination of vectors:
V = {(a, (2c - d), c, d) | c, d ∈ ℝ}
The basis of V is a set of linearly independent vectors that span V. In this case, the basis consists of {(1, 2, 0, 0), (0, -1, 0, 1)}. To check if they are linearly independent, we set up a system of equations:
a * (1, 2, 0, 0) + b * (0, -1, 0, 1) = (0, 0, 0, 0)
This leads to the following system:
a = 0
2a - b = 0
0 = 0
0 = 0
Solving this system, we find that a = 0 and b = 0. Therefore, the vectors (1, 2, 0, 0) and (0, -1, 0, 1) are linearly independent.
The basis of V is {(1, 2, 0, 0), (0, -1, 0, 1)}, and the dimension of V is 2.
b) W:
To find the basis of W, we need to find the linearly independent vectors that span the set. We can express W as a linear combination of vectors:
W = {(d, b, c, d) | b = 2c}
The basis of W is a set of linearly independent vectors that span W. In this case, the basis consists of {(1, 0, 0, 1), (0, 2, 1, 0)}. To check if they are linearly independent, we set up a system of equations:
a * (1, 0, 0, 1) + b * (0, 2, 1, 0) = (0, 0, 0, 0)
This leads to the following system:
a = 0
2b = 0
a + b = 0
a = 0
Solving this system, we find that a = 0 and b = 0. Therefore, the vectors (1, 0, 0, 1) and (0, 2, 1, 0) are linearly independent.
The basis of W is {(1, 0, 0, 1), (0, 2, 1, 0)}, and the dimension of W is 2.
c) V ∩ W:
To find the basis of V ∩ W, we need to find the vectors that are common to both V and W. From the equations given, we can see that the common condition is that a = d. Thus, we can express V ∩ W as a linear combination of vectors:
V ∩ W = {(d, b, c, d) | b = 2c, d ∈ ℝ}
Taking b = 2c in the V equation, we have:
(0, 2c, c, 0) ∈ V ∩ W
The basis of V ∩ W is a set of linearly independent vectors that span V ∩ W. In this case, the basis consists of {(0, 2, 1, 0)}. To check if it is linearly independent, we set up a system of equations:
a * (0, 2, 1, 0) = (0, 0, 0, 0)
This leads to the following system:
0 = 0
2a = 0
a = 0
0 = 0
Solving this system, we find that a = 0. Therefore, the vector (0, 2, 1, 0) is linearly independent.
The basis of V ∩ W is {(0, 2, 1, 0)}, and the dimension of V ∩ W is 1.