Linear algebra: basis of vector spaces?
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In the vector space of polynomials with real coefficients show that {5–x, 6+x, x^2+4x+1} is a basis for the subspace of polynomials with degree at most 2. Express 5x^2+7x+3 as a linear combination of these basis vectors.
Answers:
To show that {5-x, 6+x, x^2+4x+1} is a basis for the subspace of polynomials with degree at most 2, you have to show that it is a linearly independent set and that it spans the vector space of polynomials with real coefficients with degree at most 2. However, you should also notice that the dimension of the vector space of all polynomials with real coefficients with degree at most 2 is 3.
Thus, if you can show that {5-x, 6+x, x^2+4x+1} is a linearly independent set, you know that it is a maximally independent set (since the dimension of the vector space you're working with is 3, and a basis is a maximally linearly independent spanning set).
So, to show the set is linearly independent, suppose that there is the following relation:
a(5-x)+b(6+x)+c(x^2+4x+1)=0 (where 0 is the zero polynomial)
and solve for the coefficients (you want them all to be 0, as then the only solution to the above equation will be the trivial one, and hence your vectors will be linearly independent)
So, comparing coefficients, we have:
x^2: c=0
x: 4c+b-a=0. But we know c=0. Therefore b=a
x^0: c+5a+6b=0 Again we know c=0, so we have 5a+6b=0
But b=a. Therefore 11a=0 =>a=0 and therefore b=0.
Thus the only solution to the equation is the trivial one and the vectors are linearly independent, and as dim({5-x, 6+x, x^2+4x+1})=dimV=3, they form a basis.
Now to express 5x^2+7x+3 as a linear combination of these basis vectors, set up the following equation and solve it:
5x^2+7x+3= a(5-x)+b(6+x)+c(x^2+4x+1)
Just expand, collect and compare coefficients.
Good luck with your algebra!
first show that
{5–x, 6+x, x^2+4x+1} is
lineary independent
then take any arbitrary
polynomials with degree 2
and show that it can be
written as a linear combination
of given set
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In the vector space of polynomials with real coefficients show that {5–x, 6+x, x^2+4x+1} is a basis for the subspace of polynomials with degree at most 2. Express 5x^2+7x+3 as a linear combination of these basis vectors.
Answers:
To show that {5-x, 6+x, x^2+4x+1} is a basis for the subspace of polynomials with degree at most 2, you have to show that it is a linearly independent set and that it spans the vector space of polynomials with real coefficients with degree at most 2. However, you should also notice that the dimension of the vector space of all polynomials with real coefficients with degree at most 2 is 3.
Thus, if you can show that {5-x, 6+x, x^2+4x+1} is a linearly independent set, you know that it is a maximally independent set (since the dimension of the vector space you're working with is 3, and a basis is a maximally linearly independent spanning set).
So, to show the set is linearly independent, suppose that there is the following relation:
a(5-x)+b(6+x)+c(x^2+4x+1)=0 (where 0 is the zero polynomial)
and solve for the coefficients (you want them all to be 0, as then the only solution to the above equation will be the trivial one, and hence your vectors will be linearly independent)
So, comparing coefficients, we have:
x^2: c=0
x: 4c+b-a=0. But we know c=0. Therefore b=a
x^0: c+5a+6b=0 Again we know c=0, so we have 5a+6b=0
But b=a. Therefore 11a=0 =>a=0 and therefore b=0.
Thus the only solution to the equation is the trivial one and the vectors are linearly independent, and as dim({5-x, 6+x, x^2+4x+1})=dimV=3, they form a basis.
Now to express 5x^2+7x+3 as a linear combination of these basis vectors, set up the following equation and solve it:
5x^2+7x+3= a(5-x)+b(6+x)+c(x^2+4x+1)
Just expand, collect and compare coefficients.
Good luck with your algebra!
first show that
{5–x, 6+x, x^2+4x+1} is
lineary independent
then take any arbitrary
polynomials with degree 2
and show that it can be
written as a linear combination
of given set
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