Question

Determine whether each set {p1,p2} is a linearly independent set in P3.
1. The polynomials p1 (t) = 1 + t2 and p2(t) = 1 - t2.
2. The polynomials p1 (t) = 2t + t2 and p2(t) = 1 + t.
3. The polynomials p1(t) = 2t - 4t2 and p2(t) = 6t2 - 3t.

Answer 1

Answer:

1) The polynomials $p_{1}(t) = 1 +t^{2}$ and $p_{2}(t) = 1-t^{2}$ are linearly independient, 2) The polynomials $p_{1}(t) = 2\cdot t +t^{2}$ and $p_{2}(t) = 1+t$ are linearly independent, 3) The polynomials $p_{1}(t) = 2\cdot t - 4\cdot t^{2}$ and $p_{2}(t) = 6\cdot t^{2}-3\cdot t$ are linearly dependent.

Step-by-step explanation:

A set is linearly independent if and only if the sum of elements satisfy the following conditions:

$\Sigma_{i=0}^{n} \alpha_{i} \cdot u_{i} = 0$

$\alpha_{0} = \alpha_{1} =...=\alpha_{i} = 0$

1) The set of elements form the following sum:

$\alpha_{1}\cdot p_{1}(t)+\alpha_{2}\cdot p_{2}(t) = 0$

$\alpha_{1}\cdot (1+t^{2})+\alpha_{2}\cdot (1-t^{2}) = 0$

$(\alpha_{1}+\alpha_{2})\cdot (1) +(\alpha_{1}-\alpha_{2})\cdot t^{2} = 0$

From definition this system of equations must be satisfied:

$\alpha_{1} + \alpha_{2} = 0$ Eq. 1

$\alpha_{1}-\alpha_{2} = 0$ Eq. 2

From Eq. 2:

$\alpha_{1} = \alpha_{2}$

In Eq. 1:

$2\cdot \alpha_{1} =0$

$\alpha_{1} = 0$

$\alpha_{2} = 0$

The polynomials $p_{1}(t) = 1 +t^{2}$ and $p_{2}(t) = 1-t^{2}$ are linearly independient.

2) The set of elements form the following sum:

$\alpha_{1}\cdot p_{1}(t)+\alpha_{2}\cdot p_{2}(t) = 0$

$\alpha_{1}\cdot (2\cdot t+t^{2})+\alpha_{2}\cdot (1+t) = 0$

$\alpha_{2}\cdot (1) +(2\cdot \alpha_{1}+\alpha_{2})\cdot t +\alpha_1 \cdot t^{2} = 0$

From definition this system of equations must be satisfied:

$\alpha_{2} = 0$

$2\cdot \alpha_{1}+\alpha_{2} = 0$

$\alpha_{1} = 0$

The polynomials $p_{1}(t) = 2\cdot t +t^{2}$ and $p_{2}(t) = 1+t$ are linearly independent.

3) The set of elements form the following sum:

$\alpha_{1}\cdot p_{1}(t)+\alpha_{2}\cdot p_{2}(t) = 0$

$\alpha_{1}\cdot (2\cdot t-4\cdot t^{2})+\alpha_{2}\cdot (6\cdot t^{2}-3\cdot t) = 0$

$(2\cdot \alpha_{1}-3\cdot \alpha_{2})\cdot t + (-4\cdot \alpha_{1}+6\cdot \alpha_{2})\cdot t^{2} = 0$

From definition this system of equations must be satisfied:

$2\cdot \alpha_{1}-3\cdot \alpha_{2} = 0$ (Eq. 1)

$-4\cdot \alpha_{1}+6\cdot \alpha_{2} =0$ (Eq. 2)

It is easy to find that each coefficient is multiple of the other one, that is:

$\alpha_{1} =\frac{3}{2}\cdot \alpha_{2}$ (From Eq. 1)

$\alpha_{1} = \frac{6}{4}\cdot \alpha_{2}$ (From Eq. 2)

$\alpha_{1} = \frac{3}{2}\cdot \alpha_{2}$

Which means that polynomials $p_{1}(t) = 2\cdot t - 4\cdot t^{2}$ and $p_{2}(t) = 6\cdot t^{2}-3\cdot t$ are linearly dependent.