To find the basis and dimension of V and W, we need to find the linearly independent vectors that span each subspace.
a) To find a basis for V, we need to find the linearly independent solutions to the equation given. The equation for V is 4: b - 2c + d = 0. We can rewrite this equation as a linear combination of the variables:
b = 2c - d
Now, we can express this in vector form:
(a, b, c, d) = (a, 2c - d, c, d)
We can see that a, c, and d are free variables, and b can be expressed in terms of c and d. Let's solve for a, c, and d in terms of the free variables:
a = a (any value)
c = c (free variable)
d = d (free variable)
Now, we can write the vector (a, b, c, d) as a combination of the free variables:
(a, 2c - d, c, d) = a(1, 0, 0, 0) + c(0, 2, 1, 0) + d(0, -1, 0, 1)
So, the basis for V is {(1, 0, 0, 0), (0, 2, 1, 0), (0, -1, 0, 1)}. The dimension of V is the number of vectors in its basis, which is 3.
b) To find a basis for W, we need to find the linearly independent solutions to the equation given. The equation for W is 4: a = d and b = 2c. We can express this in vector form:
(a, b, c, d) = (d, 2c, c, d)
We can see that c and d are free variables, and a and b can be expressed in terms of c and d. Let's solve for c and d in terms of the free variables:
c = c (free variable)
d = d (free variable)
Now, we can write the vector (a, b, c, d) as a combination of the free variables:
(d, 2c, c, d) = c(0, 2, 1, 0) + d(1, 0, 0, 1)
So, the basis for W is {(0, 2, 1, 0), (1, 0, 0, 1)}. The dimension of W is the number of vectors in its basis, which is 2.
c) Since a) and b) give the basis and dimension of V and W respectively, we don't need to find a basis or dimension for the entire space.
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)
1 answer