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The linear span

Polynomials[edit]

Let K be RC, or any field, and let V be the set P of all polynomials with coefficients taken from the field K. Consider the vectors (polynomials) p1 := 1, p2 := x + 1, and p3 := x2 + x + 1.

Is the polynomial x2 − 1 a linear combination of p1p2, and p3? To find out, consider an arbitrary linear combination of these vectors and try to see when it equals the desired vector x2 − 1. Picking arbitrary coefficients a1a2, and a3, we want{\displaystyle a_{1}(1)+a_{2}(x+1)+a_{3}(x^{2}+x+1)=x^{2}-1.\,}

a_{1}(1)+a_{2}(x+1)+a_{3}(x^{2}+x+1)=x^{2}-1.\,

Multiplying the polynomials out, this means{\displaystyle (a_{1})+(a_{2}x+a_{2})+(a_{3}x^{2}+a_{3}x+a_{3})=x^{2}-1\,}

(a_{1})+(a_{2}x+a_{2})+(a_{3}x^{2}+a_{3}x+a_{3})=x^{2}-1\,

and collecting like powers of x, we get{\displaystyle a_{3}x^{2}+(a_{2}+a_{3})x+(a_{1}+a_{2}+a_{3})=1x^{2}+0x+(-1).\,}

a_{3}x^{2}+(a_{2}+a_{3})x+(a_{1}+a_{2}+a_{3})=1x^{2}+0x+(-1).\,

Two polynomials are equal if and only if their corresponding coefficients are equal, so we can conclude{\displaystyle a_{3}=1,\quad a_{2}+a_{3}=0,\quad a_{1}+a_{2}+a_{3}=-1.\,}

a_{3}=1,\quad a_{2}+a_{3}=0,\quad a_{1}+a_{2}+a_{3}=-1.\,

This system of linear equations can easily be solved. First, the first equation simply says that a3 is 1. Knowing that, we can solve the second equation for a2, which comes out to −1. Finally, the last equation tells us that a1 is also −1. Therefore, the only possible way to get a linear combination is with these coefficients. Indeed,{\displaystyle x^{2}-1=-1-(x+1)+(x^{2}+x+1)=-p_{1}-p_{2}+p_{3}\,}

x^{2}-1=-1-(x+1)+(x^{2}+x+1)=-p_{1}-p_{2}+p_{3}\,

so x2 − 1 is a linear combination of p1p2, and p3.

On the other hand, what about the polynomial x3 − 1? If we try to make this vector a linear combination of p1p2, and p3, then following the same process as before, we’ll get the equation{\displaystyle 0x^{3}+a_{3}x^{2}+(a_{2}+a_{3})x+(a_{1}+a_{2}+a_{3})\,}

0x^{3}+a_{3}x^{2}+(a_{2}+a_{3})x+(a_{1}+a_{2}+a_{3})\,

{\displaystyle =1x^{3}+0x^{2}+0x+(-1).\,}

=1x^{3}+0x^{2}+0x+(-1).\,

However, when we set corresponding coefficients equal in this case, the equation for x3 is{\displaystyle 0=1\,}

0=1\,

which is always false. Therefore, there is no way for this to work, and x3 − 1 is not a linear combination of p1p2, and p3.