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.
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.
1 answer