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:Part 1:Is the set of polynomials <math> x^2, x, 1 </math> a basis for the set of all polynomials of degree two or less? | :Part 1:Is the set of polynomials <math> x^2, x, 1 </math> a basis for the set of all polynomials of degree two or less? | ||
| − | + | :solution: | |
YES. The first way to approach this problem is to write the knowns. We know that our given subspace V is the set of all polynomials of degree two or less. We also know that <math> x^2, x, 1 </math> are the vectors we'd like to test as to whether they form a basis for our space V. There are two conditions, <math> x^2, x, 1 </math> need to be linearly independent and <math> x^2, x, 1 </math> need to span V. To be concise, let's call these three vectors, respectively, as <math> \vec v_1, \vec v_2, \vec v_3 </math> | YES. The first way to approach this problem is to write the knowns. We know that our given subspace V is the set of all polynomials of degree two or less. We also know that <math> x^2, x, 1 </math> are the vectors we'd like to test as to whether they form a basis for our space V. There are two conditions, <math> x^2, x, 1 </math> need to be linearly independent and <math> x^2, x, 1 </math> need to span V. To be concise, let's call these three vectors, respectively, as <math> \vec v_1, \vec v_2, \vec v_3 </math> | ||
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The first aspect is trivial, since you can't make any one term by combining the others. | The first aspect is trivial, since you can't make any one term by combining the others. | ||
| − | As for the second part, here's a simple trick: every polynomial (let's just call it p(x)) can be written in the form <math> p(x)=a*x^2+ b*x+ c*1 </math>. However, if we look more carefully, a, b, and c are all constants! So we can rewrite this as <math> p(x)=c_1*x^2+ c_2*x+ c_3*1 </math>. This however can be also rewritten as <math> p(x)=c_1*\vec v_1+ c_2*\vec v_2 | + | |
| + | As for the second part, here's a simple trick: every polynomial (let's just call it p(x)) can be written in the form <math> p(x)=a*x^2+ b*x+ c*1 </math>. However, if we look more carefully, a, b, and c are all constants! So we can rewrite this as <math> p(x)=c_1*x^2+ c_2*x+ c_3*1 </math>. This however can be also rewritten as <math> p(x)=c_1*\vec v_1+ c_2*\vec v_2+ c_3*\vec v_3 </math> and notice something? this is the definition of our three vectors spanning the space! Therefore the three vectors do indeed span the space and are linearly independent. This means that they are a basis | ||
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As a take away from this problem, notice that to test whether the basis vectors span the space, write the ambiguous form of the polynomial (in this case, <math> p(x)=a*x^2+ b*x+ c*1 </math>) and then manipulate it to see if it can be written as an arbitrary linear combination of our given basis vectors. This is how you test for span. NOTE: It wasn't absolutely necessary to rewrite the basis vectors in the from <math>\vec v_1, \vec v_2, \vec v_3 </math>, but this was to help make the span more familiar. | As a take away from this problem, notice that to test whether the basis vectors span the space, write the ambiguous form of the polynomial (in this case, <math> p(x)=a*x^2+ b*x+ c*1 </math>) and then manipulate it to see if it can be written as an arbitrary linear combination of our given basis vectors. This is how you test for span. NOTE: It wasn't absolutely necessary to rewrite the basis vectors in the from <math>\vec v_1, \vec v_2, \vec v_3 </math>, but this was to help make the span more familiar. | ||
| + | |||
:Part 2: is the set of polynomials <math> 3x^2, x ,1 </math> a basis for the set of all polynomials of degree two or less? | :Part 2: is the set of polynomials <math> 3x^2, x ,1 </math> a basis for the set of all polynomials of degree two or less? | ||
| − | + | :solution: | |
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| + | YES. Similar to Part 1, we have the same space V and we know that the vectors are linearly independent. Moreover, when we write <math> p(x)=c_1*x^2+ c_2*x+ c_3*1 </math>, we know that <math>3x^2=3*(x^2)</math> so <math> p(x)=(c_1 /3)*(3x^2)+ c_2*x+ c_3*1 </math> and therefore since <math> c_1 /3</math> is a constant, we can rewrite this all as <math> p(x)=(k_1)*(3x^2)+ k_2*x+ k_3*1 </math>. Which means the set of vectors span the space. | ||
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| + | Another way of seeing this is that this basis can form the previous basis, by simply dividing the first vector by three. Therefore, since the previous basis spanned the space, this one must too. That would save a lot of work. | ||
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:Part 3: is the set of polynomials <math> 3x^2 + x, x , 1 </math> a basis for the set of all polynomials of degree two or less? | :Part 3: is the set of polynomials <math> 3x^2 + x, x , 1 </math> a basis for the set of all polynomials of degree two or less? | ||
Revision as of 08:37, 12 March 2013
Basis Problems
POLYNOMIALS
Example #1: Testing whether a set of vectors is a basis for a given space
- Part 1:Is the set of polynomials $ x^2, x, 1 $ a basis for the set of all polynomials of degree two or less?
- solution:
YES. The first way to approach this problem is to write the knowns. We know that our given subspace V is the set of all polynomials of degree two or less. We also know that $ x^2, x, 1 $ are the vectors we'd like to test as to whether they form a basis for our space V. There are two conditions, $ x^2, x, 1 $ need to be linearly independent and $ x^2, x, 1 $ need to span V. To be concise, let's call these three vectors, respectively, as $ \vec v_1, \vec v_2, \vec v_3 $
The first aspect is trivial, since you can't make any one term by combining the others.
As for the second part, here's a simple trick: every polynomial (let's just call it p(x)) can be written in the form $ p(x)=a*x^2+ b*x+ c*1 $. However, if we look more carefully, a, b, and c are all constants! So we can rewrite this as $ p(x)=c_1*x^2+ c_2*x+ c_3*1 $. This however can be also rewritten as $ p(x)=c_1*\vec v_1+ c_2*\vec v_2+ c_3*\vec v_3 $ and notice something? this is the definition of our three vectors spanning the space! Therefore the three vectors do indeed span the space and are linearly independent. This means that they are a basis
As a take away from this problem, notice that to test whether the basis vectors span the space, write the ambiguous form of the polynomial (in this case, $ p(x)=a*x^2+ b*x+ c*1 $) and then manipulate it to see if it can be written as an arbitrary linear combination of our given basis vectors. This is how you test for span. NOTE: It wasn't absolutely necessary to rewrite the basis vectors in the from $ \vec v_1, \vec v_2, \vec v_3 $, but this was to help make the span more familiar.
- Part 2: is the set of polynomials $ 3x^2, x ,1 $ a basis for the set of all polynomials of degree two or less?
- solution:
YES. Similar to Part 1, we have the same space V and we know that the vectors are linearly independent. Moreover, when we write $ p(x)=c_1*x^2+ c_2*x+ c_3*1 $, we know that $ 3x^2=3*(x^2) $ so $ p(x)=(c_1 /3)*(3x^2)+ c_2*x+ c_3*1 $ and therefore since $ c_1 /3 $ is a constant, we can rewrite this all as $ p(x)=(k_1)*(3x^2)+ k_2*x+ k_3*1 $. Which means the set of vectors span the space.
Another way of seeing this is that this basis can form the previous basis, by simply dividing the first vector by three. Therefore, since the previous basis spanned the space, this one must too. That would save a lot of work.
- Part 3: is the set of polynomials $ 3x^2 + x, x , 1 $ a basis for the set of all polynomials of degree two or less?
- quick solution:
- rigorous solution:
- Part 4: is the set of polynomials $ 3x^2+x+1, 2x+1, , 2 $ a basis for the set of all polynomials of degree two or less?
- quick solution:
- rigorous solution:
- Part 5: is the set of polynomials $ x^3, x, , 1 $ a basis for the set of all polynomials of degree two or less?
- quick solution: No. x^3 is included in the space of polynomials of degree THREE. So this set of vectors cannot be a basis for polynomials of degree two or less since its span is not equal to the given space of polynomials.
- Part 6: is the set of polynomials $ x^2,3x^2, x , 1 $ a basis for the set of all polynomials of degree two or less?
- quick solution:
- rigorous solution:
- Part 7: is the set of polynomials $ x^2,3x^2 + 1, x , 1 $ a basis for the set of all polynomials of degree two or less?
- quick solution:
- rigorous solution:
- Part 8: is the set of polynomials $ x^2, x , 1 $ a basis for the set of all polynomials of degree THREE or less?
- quick solution:
- rigorous solution:
- Part 9: is the set of polynomials $ x^2, x , 1 $ a basis for the set of all polynomials of degree ONE or less?
- quick solution:
- rigorous solution:
- Part 10: is the set of polynomials $ x^2, x , 1 $ a basis for the set of all polynomials of EXACTLY degree TWO?
- solution: NO, it is not a basis for the set of all polynomials of exactly degree two. $ 0*x^2+0*x+0*1=0 $ 0 is not a polynomial of degree two. So this set of polynomials spans outside of the given space of polynomials.
- Part 11: is it possible to make a basis for the set of all polynomials of EXACTLY degree TWO?
- solution: Nope. Every basis can make the $ \vec 0 $ by having constants of 0 in front of the basis vectors and summing them. In other words, the span of the basis vectors always contains 0 and 0 is not a polynomial of exactly degree two. In fact, it is impossible to make a basis for polynomials of and Exact degree greater than 0, since none of those spaces contain zero.
- Part 12: is the set of polynomials $ x^2, x , 1 $ a basis for the set of ALL POLYNOMIALS?
- quick solution: Nope. All polynomials includes polynomials of degree three or higher. As seen from Part 8, this set of vectors isn't a basis for this space.
- Part 13: Is it possible to make a finite basis for the set of ALL POLYNOMIALS?
- solution: Nope. To summarize the proof, by having a finite basis, we will have polynomials of at most degree n. But All polynomials includes polynomials of degree n+1, so the basis won't suffice.
Example #2: Finding the space to which a set of polynomials corresponds to. Example #3: Finding a basis of a given space.
