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)
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.
To find the basis and dimension of a vector space, we need to determine a set of linearly independent vectors that span the space.
a) To find the basis and dimension of V, we need to solve the equation given and express it in terms of free variables.
From the equation in V: b - 2c + d = 0, we can write it as:
b = 2c - d
Now, let's express the basis vectors of V in terms of free variables:
v = (a, b, c, d) = (a, 2c - d, c, d)
There are 4 variables (a, b, c, d) and 2 equations in V. It means there will be 4 - 2 = 2 free variables.
Let's express the basis vectors using the free variables:
v = (a, 2c - d, c, d) = a(1, 0, 0, 0) + (0, 2, 1, 0) + (0, -1, 0, 1)
Therefore, a 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 the basis, which is 3.
b) To find the basis and dimension of W, we need to solve the equation given and express it in terms of free variables.
From the equation in W: a = d and b = 2c, we can write it as:
d = a
b = 2c
Now, let's express the basis vectors of W in terms of free variables:
w = (a, b, c, d) = (a, 2c, c, a)
There are 4 variables (a, b, c, d) and 2 equations in W. It means there will be 4 - 2 = 2 free variables.
Let's express the basis vectors using the free variables:
w = (a, 2c, c, a) = a(1, 0, 0, 1) + (0, 2, 1, 0) + (0, 0, 1, 0)
Therefore, a basis for W is {(1, 0, 0, 1), (0, 2, 1, 0), (0, 0, 1, 0)}.
The dimension of W is the number of vectors in the basis, which is 3.
c) The intersection of V and W is the set of vectors that satisfy both the equations given for V and W.
Equation 1: b - 2c + d = 0
Equation 2: a = d, b = 2c
To find the intersection, we can substitute the values from Equation 2 into Equation 1:
(2c - 2c) + a = 0
a = 0
So, the vectors in the intersection space will have a = 0.
The basis for the intersection space is {(0, 0, c, 0)}.
The dimension of the intersection space is 1, as it has only one vector in the basis.