Let W be the space generated by the polynomials x3 + 3x2 – x + 4, and
2x3 + x2 – 7x – 7. Find a basis and the dimension of W.
To find a basis for W, we need to determine the linearly independent polynomials that generate the space.
First, let's rewrite the given polynomials:
P1(x) = x^3 + 3x^2 - x + 4
P2(x) = 2x^3 + x^2 - 7x - 7
We can see that both polynomials have degree 3, so the basis of W must include at least one polynomial of degree 3. Let's start by considering P1(x):
P1(x) = x^3 + 3x^2 - x + 4 = 1*x^3 + 3*x^2 - 1*x + 4
We can write P1(x) as a linear combination of the powers of x:
P1(x) = 1*x^3 + 3*x^2 - 1*x + 4 = 1*x^3 + 3*x^2 + (-1)*x + 4*x^0
So, P1(x) can be generated by the monomials {x^3, x^2, x^0}. These monomials are linearly independent since none can be expressed as a combination of the others.
Now let's consider P2(x):
P2(x) = 2*x^3 + x^2 - 7*x - 7 = 2*x^3 + 1*x^2 + (-7)*x + (-7)*x^0
So, P2(x) can be generated by the monomials {x^3, x^2, x^0}. These monomials are also linearly independent.
Since both polynomials can be generated by the same set of linearly independent monomials, we can conclude that any other polynomial in the space W can also be generated by these monomials. Therefore, a basis for W is {x^3, x^2, x^0}.
The dimension of a vector space is equal to the number of elements in its basis. In this case, the dimension of W is 3, as the basis {x^3, x^2, x^0} has 3 elements.
To find a basis for the space W, we need to find a linearly independent set of polynomials that can generate the space W. We can do this by reducing the given polynomials to row-echelon form.
First, let's set up an augmented matrix of the given polynomials:
[1 3 -1 4 | 0]
[2 1 -7 -7 | 0]
Using Gaussian elimination, we can reduce the matrix to row-echelon form:
[1 3 -1 4 | 0]
[0 -5 -5 -15 | 0]
Now, let's write the row-echelon form as equations:
x1 + 3x2 - x3 + 4x4 = 0
-5x2 - 5x3 - 15x4 = 0
From this, we can see that the variables x2, x3, and x4 are free variables. We can solve for x1 in terms of these free variables:
x1 = -3x2 + x3 - 4x4
A basis for W can be found by letting the free variables be equal to 1 and the rest equal to 0. Let's take x2 = 1, x3 = 0, x4 = 0:
x1 = -3(1) + 0 - 4(0) = -3
So, one basis vector is -3.
Let's take x2 = 0, x3 = 1, x4 = 0:
x1 = -3(0) + 1 - 4(0) = 1
So, another basis vector is 1.
Finally, let's take x2 = 0, x3 = 0, x4 = 1:
x1 = -3(0) + 0 - 4(1) = -4
So, a third basis vector is -4.
Therefore, a basis for W is {-3, 1, -4}. The dimension of W is the number of basis vectors, which in this case is 3.