Determining a Basis for P2: Alpha Values

[snoggerT]

Determine all values of the constant (alpha) for which
{1+(alpha)*x^2, 1+x+x^2, 2 +x} is a basis for P2

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?

 

[jhicks]

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]

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]

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

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 $v_1 ,v_2 ,v_3$ are linearly independent. This means that we want to find where $v_1 \neq c_1 v_2 + c_2 v_3$, where $c_1$ and $c_2$ are constants. This means we can formulate 3 equations corresponding to the coefficients on the powers of x for $v_1$ to find where they *are* linearly dependent and not include the corresponding value(s) of $\alpha$ in the solution:

$1 = c_1 + 2c_2$
$0 = c_1 + c_2$
$\alpha = c_1$

Solving for $c_1$ and $c_2$ we find that $c_1 = -1$ and $c_2 = 1$ corresponding to an $\alpha = -1$.

Edit: $$\LaTeX$$ is showing up a bit weird so maybe check back in a few.

 

[HallsofIvy]

Determine all values of the constant (alpha) for which
{1+(alpha)*x^2, 1+x+x^2, 2 +x} is a basis for P2

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

2) For what values of $\alpha$ are $1+ \alpha x^2$, $1+ x+ x^2$, and 2+ x independent?

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

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

2) For what values of $\alpha$ are $1+ \alpha x^2$, $1+ x+ x^2$, and 2+ x independent?

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