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Given the polynomias $$q_1(x)=2x+1 \\ q_2(x)=4x+1\\ q_3(x)=x^3+x^2+x+1\\ q_4(x)=x^3+x-2 $$

is it possible to write $q_5(x)=2x^2+3x+4$ as a linear combination of $q_1,q_2,q_3,q_4$?

How do I tackle this?

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    $\begingroup$ It is unclear from the question what coefficients are allowed in forming a linear combination. It matters whether coefficients are integers or rational numbers or polynomials (whether with integer or rational coefficients). I put an answer for integers in case this is what you are looking for - the others assume rationals. $\endgroup$ Jul 4, 2013 at 18:04
  • $\begingroup$ We gotta have $2$ of $q_2$ because it's the only one with a quadratic term. So we need $-2$ of $q_4$ to cancel out the cubic term. Now just see if there's a linear combination of $q_1$ and $q_2$ to take care of the linear term and the constant. $\endgroup$
    – Jack M
    Jul 4, 2013 at 20:38

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Hint: $$1=2q_1-q_2$$ $$x=\frac{1}{2}q_2-\frac{1}{2}q_1$$ $$x^2=q_3-q_4-3=q_3-q_4+3q_2-6q_1$$

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While writing the answer I stumbled upon a part in my lecture notes that answered my question. Maybe this can be of help to others.

You need to prove, that $$2x^2+3x+4 = a\cdot q_1 + b \cdot q_2 + c \cdot q_3 + d \cdot q_4$$

To do this, factorise the expression on the right side with the polynomials filled in:

$$2x^2+3x+4 = x^3(c+d)+x^2(c)+x(2a+4b+c+d)+a+b+c-2d$$ Now we get a linear system of equations:

$$0=c+d\tag{1}$$ $$2=c\tag{2}$$ $$3=2a+4b+c+d\tag{3}$$ $$4=a+b+c-2d\tag{4}$$

By solving this, you will either get the scalars for a linear combination of the given polynomials to represent $q_5$ or evidence, that there's no such combination.

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  • $\begingroup$ This approach will work in general, but in this example there's a simpler way. $\endgroup$
    – vadim123
    Jul 4, 2013 at 17:15
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Well, there's an easier (imho) way. Let's see if we can produce a canonical basis in $P_3[x]$ by using $q_{1..4}$.

Linear combinations of $q_1$ and $q_2$ allow us to generate polynomials $x$ and $1$, thus we can replace $q_{1..4}$ by polynomials $1$, $x$, $x^3+x^2$, $x^3$. Clearly, linear combinations of two last polynomials generate $x^2$ and $x^3$, thus $q_{1..4}$ are linearly independent and are the basis in $P_3[x]$ (because there're 4 of them and they generate $1$, $x$, $x^2$, $x^3$), hence $q_5$ is indeed a linear combination of $q_{1..4}$.

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  • $\begingroup$ That seems to be what I was about to type. Seems like a lot easier than what's done before since you don't need to make any horrible calculations. (I would change 'the basis' to 'the standard basis' though) $\endgroup$
    – HSN
    Jul 4, 2013 at 17:17
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If we are looking at a linear combination with integer coefficients, let's check the parity of the coefficient of $x$.

In the first two polynomials this is even.

In the third and fourth the coefficient of $x$ is the same as the coefficient of $x^3$. To eliminate $x^3$, therefore, we have to reduce the contribution to the coefficient of $x$ from these two polynomials to zero too.

Therefore any $\mathbb Z-$linear combination of our given basis polynomials which has zero coefficient for $x^3$ has an even coefficient for $x$. Since this is not the case for the target polynomial, we can't make it.

Note that $($coefficient of $x^3+$ coefficient of $x)$ is always even, which is a more general parity statement.

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