# Determining a Basis for P2: Alpha Values

• snoggerT
In summary, we need to find all values of alpha such that the set of vectors {1+(alpha)*x^2, 1+x+x^2, 2 +x} forms a basis for P2. This can be done by checking for linear independence or by checking if the set spans the space of all quadratic polynomials. Solving for both cases, we find that alpha = -1 is the only value that satisfies both conditions, making it the only possible value for alpha. This problem has a simple solution and can be solved without using a matrix of coefficients.
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?

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.

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?

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.

Last edited:
snoggerT said:
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!

HallsofIvy said:
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.

## 1. What is the purpose of determining a basis for P2: Alpha Values?

The purpose of determining a basis for P2: Alpha Values is to identify the minimum set of independent variables that can accurately represent the data in a linear regression model. This helps to simplify the model without sacrificing its predictive power.

## 2. How do you determine the basis for P2: Alpha Values?

The basis for P2: Alpha Values can be determined by using a variety of statistical methods, such as stepwise regression or forward/backward selection. These methods involve systematically adding or removing variables from the model to find the best combination of predictors.

## 3. What factors should be considered when determining a basis for P2: Alpha Values?

When determining a basis for P2: Alpha Values, it is important to consider the significance and relevance of each variable to the outcome being predicted, as well as the potential for multicollinearity (high correlation between variables). Additionally, the chosen basis should be simple and easily interpretable.

## 4. Can the basis for P2: Alpha Values change over time?

Yes, the basis for P2: Alpha Values can change over time as the data being analyzed may change or new variables may become available. It is important to regularly review and update the basis to ensure the model remains accurate and relevant.

## 5. What are the potential limitations of determining a basis for P2: Alpha Values?

The process of determining a basis for P2: Alpha Values is not foolproof and may result in a model that is not completely accurate. Additionally, there may be variables that are not included in the analysis that could impact the outcome being predicted. It is important to carefully consider the chosen basis and continuously evaluate the model's performance.

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