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Determining a basis

by snoggerT
Tags: basis, determining
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snoggerT
#1
Mar3-08, 10:38 PM
P: 186
Determine all values of the constant (alpha) for which
{1+(alpha)*x^2, 1+x+x^2, 2 +x} is a basis for P2





3. The attempt at a solution

once again I'm not real sure how to solve this problem. Would I set it up as a system of linear equations? How do you start a problem like this?
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jhicks
#2
Mar3-08, 10:50 PM
P: 336
Well the basis is the smallest set of vectors that spans a space. It might be easiest to determine all values of alpha such that those 3 vectors are not linearly independent. Consider, for example, alpha = -1.
snoggerT
#3
Mar3-08, 10:53 PM
P: 186
Can you explain how linear independence works on a vector like that? I know that -1 is the answer, but why wouldn't say -2 or any other negative number?

jhicks
#4
Mar3-08, 11:02 PM
P: 336
Determining a basis

So you have 3 potential basis vectors: [itex]v_1 = 1+\alpha x^2[/itex], [itex]v_2 = 1+x+x^2[/itex] and [itex]v_3 = 2+x[/itex]

Since there are 3 vectors and you won't find a basis of P2 that has less than 3 members, it suffices to find where [itex]v_1 ,v_2 ,v_3[/itex] are linearly independent. This means that we want to find where [itex]v_1 \neq c_1 v_2 + c_2 v_3[/itex], where [itex]c_1[/itex] and [itex]c_2[/itex] are constants. This means we can formulate 3 equations corresponding to the coefficients on the powers of x for [itex]v_1[/itex] to find where they *are* linearly dependent and not include the corresponding value(s) of [itex]\alpha[/itex] in the solution:

[itex]1 = c_1 + 2c_2[/itex]
[itex]0 = c_1 + c_2[/itex]
[itex]\alpha = c_1[/itex]

Solving for [itex]c_1[/itex] and [itex]c_2[/itex] we find that [itex]c_1 = -1[/itex] and [itex]c_2 = 1[/itex] corresponding to an [itex]\alpha = -1[/itex].

Edit: [tex]\LaTeX[/tex] is showing up a bit weird so maybe check back in a few.
HallsofIvy
#5
Mar4-08, 06:39 AM
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Quote Quote by snoggerT View Post
Determine all values of the constant (alpha) for which
{1+(alpha)*x^2, 1+x+x^2, 2 +x} is a basis for P2





3. The attempt at a solution

once again I'm not real sure how to solve this problem. Would I set it up as a system of linear equations? How do you start a problem like this?
You know, I guess, that P2, the space of all quadratic polynomials, has dimension 3. Knowing that you can show a set of 3 vectors has is a basis by showing either:
1) it spans the space or
2) its vectors are independent.

1) For what values of [itex]\alpha[/itex] do [itex]1+ \alpha x^2[/itex], [itex]1+ x+ x^2[/itex], and 2+ x span all of P2?
Any "vector" in P2 can be written as [itex]a^2+ bx+ c[/itex]. We need to find p, q, r so that [itex]p(1+ \alpha x^2)+ q(1+ x+ x^2)+ r(2+ x)= ax^2+ bx+ c[/itex]. Multiplying out the right side and collecting coefficients of like powers, [itex](p\alpha+ q)x^2+ (q+ r)x+ (p+ q+ 2r)= ax^2+ bc+ c[/itex]. For that to be true for all x, corresp[onding coefficients must be the same: [itex]p\alpha+ q= a, q+ r= b, p+ q+ r= c[/itex]. For what values of [itex]\alpha[/itex] does that systme have a solution? (It might be simpler to ask, "for what values of [itex]\alpha[/itex] does that not have a solution.

2) For what values of [itex]\alpha[/itex] are [itex]1+ \alpha x^2[/itex], [itex]1+ x+ x^2[/itex], and 2+ x independent?

They will be independent if the only way we can have [itex]p(1+ \alpha x^2)+ q(1+ x+ x^2)+ r(2+ x)= 0[/itex] is to have p= q= r= 0. Again, multiplying out the left side and collecting coefficients of like powers, that is [itex](p\alpha+ q)x^2+ (q+ r)x+ (p+ q+ 2r)= 0[/itex]. For that to be true for all x, all coefficients must be 0: [itex]p\alpha+ q= 0, q+ r= 0, p+ q+ r= 0[/itex]. For what values of [itex]\alpha[/itex] is the only solution to that system of equations r= p= q= 0?

Notice that in (2) you have the same system of equations as in 1 except they are equal to 0 instead of the arbitrary numbers a, b, c. Typically, it is easier to prove "independence" rather than "spanning".

And, by the way, this problem has a very simple answer!
snoggerT
#6
Mar4-08, 11:28 AM
P: 186
Quote Quote by HallsofIvy View Post
You know, I guess, that P2, the space of all quadratic polynomials, has dimension 3. Knowing that you can show a set of 3 vectors has is a basis by showing either:
1) it spans the space or
2) its vectors are independent.

1) For what values of [itex]\alpha[/itex] do [itex]1+ \alpha x^2[/itex], [itex]1+ x+ x^2[/itex], and 2+ x span all of P2?
Any "vector" in P2 can be written as [itex]a^2+ bx+ c[/itex]. We need to find p, q, r so that [itex]p(1+ \alpha x^2)+ q(1+ x+ x^2)+ r(2+ x)= ax^2+ bx+ c[/itex]. Multiplying out the right side and collecting coefficients of like powers, [itex](p\alpha+ q)x^2+ (q+ r)x+ (p+ q+ 2r)= ax^2+ bc+ c[/itex]. For that to be true for all x, corresp[onding coefficients must be the same: [itex]p\alpha+ q= a, q+ r= b, p+ q+ r= c[/itex]. For what values of [itex]\alpha[/itex] does that systme have a solution? (It might be simpler to ask, "for what values of [itex]\alpha[/itex] does that not have a solution.

2) For what values of [itex]\alpha[/itex] are [itex]1+ \alpha x^2[/itex], [itex]1+ x+ x^2[/itex], and 2+ x independent?

They will be independent if the only way we can have [itex]p(1+ \alpha x^2)+ q(1+ x+ x^2)+ r(2+ x)= 0[/itex] is to have p= q= r= 0. Again, multiplying out the left side and collecting coefficients of like powers, that is [itex](p\alpha+ q)x^2+ (q+ r)x+ (p+ q+ 2r)= 0[/itex]. For that to be true for all x, all coefficients must be 0: [itex]p\alpha+ q= 0, q+ r= 0, p+ q+ r= 0[/itex]. For what values of [itex]\alpha[/itex] is the only solution to that system of equations r= p= q= 0?

Notice that in (2) you have the same system of equations as in 1 except they are equal to 0 instead of the arbitrary numbers a, b, c. Typically, it is easier to prove "independence" rather than "spanning".

And, by the way, this problem has a very simple answer!
- What happened to the 2 in the (p+q+2r) portion of the equation when you set it to 0?

Also, would you solve using a matrix of coefficients? If so, how would you use alpha in it? I understand how you set it up, but not how to solve it.


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