# Lesson: Span and Linear Independence in Polynomials

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## Transcript

Just as we did with Rn and matrices, we can define spanning sets and linear independence of polynomials as well.

So let B be the set {p1(x) through pk(x)} be a set of polynomials of degree at most n. Then the span of B is defined as Span B = the set of all t1p1(x) + dot dot dot + tkpk(x) for scalars t1 through tk in R. So the span of B is the set of all linear combinations of our polynomials p1(x) through pk(x).

For example, let’s let B be the set {1 + x + x-squared, 1 + 2x + 3(x-squared), -5 – 5(x-squared)}. Well, then the polynomial -24 + 8x – 20(x-squared) is in the span of B since we can write it as 4(1 + x + x-squared) + 2(1 + 2x + 3(x-squared)) + 6(-5x – 5(x-squared)). We also have that the polynomial x-squared is in the span of B since (-1)(1 + x + x-squared) + (1/2)(1 + 2x + 3(x-squared)) – (1/10)(-5 – 5(x-squared)) = x-squared.

To see if the polynomial 6 + x + 6(x-squared) is in the span of B, we need to try to find scalars t1, t2, and t3 that satisfy the equation t1(1 + x + x-squared) + t2(1 + 2x + 3(x-squared)) + t3(-5 – 5(x-squared)) equals our polynomial 6 + x + 6(x-squared). Performing the calculation on the left, we see that this is the same as looking to see if 6 + x + 6(x-squared) = (t1 + t2 – 5t3) + (t1 + 2t2)x + (t1 + 3t2 – 5t3)(x-squared). Setting the coefficients equal to each other, we find that we are looking for a solution to the following system of equations. Looking at our ones, t1 + t2 – 5t3 = 6. Looking at our x terms, t1 + 2t2 = 1. And from our x-squared terms, we get t1 + 3t2 – 5t3 = 6. To solve this system, we will row reduce the augmented matrix for the system, as seen here. From our final matrix, we see that the solution to the system is that t1 = 1, t2 = 0, and t3 = -1. So this means that we can write 6 + x + 6x-squared as (1 + x + x-squared) + 0(1 + 2x + 3(x-squared) – (-5 – 5(x-squared)), and that, yes, 6 + x + 6(x-squared) is in the span of B.

As usual, after we define span, we define linear independence. The set B = {p1 through pk(x)} is said to be linearly independent if the only solution to the equation t1p1(x) through tkpk(x) = 0 is the trivial solution t1 through tk all equal to 0. Otherwise, B is said to be linearly dependent.

For example, to determine whether or not the set B is linearly independent, where B equals this, we need to look for a non-trivial solution to the equation t1(1 + 2x + 3(x-squared) + 4(x-cubed)) + t2(2 – x + 7(x-squared) + 5(x-cubed)) + t3(-4 – 4x – 8(x-squared) – 8(x-cubed)) + t4(-2 – x + 3(x-squared) + 3(x-cubed)) = 0. Performing the calculation on the left, we first distribute our scalars, and then line it up into columns to see that the linear combination becomes the single polynomial (t1 + 2t2 – 4t3 – 2t4) + (2t1 – t2 – 4t3 – t4)x + (3t1 + 7t2 – 8t3 + 3t4)(x-squared) + (4t1 + 5t2 – 8t3 + 3t4)(x-cubed), and we’re trying to set this equal to 0 + 0x + 0(x-squared) + 0(x-cubed). We want to set the coefficients equal to each other to develop the following homogeneous system of linear equations. And of course, we solve the system of homogeneous equations by row reducing its coefficient matrix, as seen here. Our last matrix is in row echelon form, and from it, we see that the rank of the coefficient matrix is 4. Now, since this is equal to the number of variables, our system has only one solution—specifically, the trivial solution—and this means that our set B is linearly independent.