To determine if each set $\{ p_1, p_2 \}$ is linearly independent in $\mathbb{P}_3$ (the space of polynomials of degree at most 3), we check if there exist constants $c_1$ and $c_2$, not both zero, such that:

$c_1 p_1(t) + c_2 p_2(t) = 0$

If the only solution to this equation is $c_1 = 0$ and $c_2 = 0$, then $\{ p_1, p_2 \}$ is linearly independent. Otherwise, they are linearly dependent.

Let’s analyze each pair:

1. Polynomials $p_1(t) = 1 + t^2$ and $p_2(t) = 1 - t^2$

To check for linear independence, set up the equation:

$c_1 (1 + t^2) + c_2 (1 - t^2) = 0$

Expanding and combining terms gives:

$(c_1 + c_2) + (c_1 - c_2) t^2 = 0$

For this equation to hold for all $t$, each coefficient must be zero:

1. $c_1 + c_2 = 0$
2. $c_1 - c_2 = 0$

Solving this system:

From equation (2), $c_1 = c_2$. Substituting $c_2 = c_1$ in equation (1):

$c_1 + c_1 = 0 \Rightarrow 2c_1 = 0 \Rightarrow c_1 = 0 \text{ and } c_2 = 0$

Since the only solution is $c_1 = 0$ and $c_2 = 0$, these polynomials are linearly independent. Answer: Yes.

2. Polynomials $p_1(t) = 2t + t^2$ and $p_2(t) = 1 + t$

Set up the equation:

$c_1 (2t + t^2) + c_2 (1 + t) = 0$

Expanding and combining terms:

$c_2 + (2c_1 + c_2)t + c_1 t^2 = 0$

Equating each coefficient to zero:

1. $c_2 = 0$
2. $2c_1 + c_2 = 0$
3. $c_1 = 0$

From equation (3), $c_1 = 0$. Substituting $c_1 = 0$ into equation (2):

$2(0) + c_2 = 0 \Rightarrow c_2 = 0$

Since the only solution is $c_1 = 0$ and $c_2 = 0$, these polynomials are linearly independent. Answer: Yes.

3. Polynomials $p_1(t) = 2t - 4t^2$ and $p_2(t) = 6t^2 - 3t$

Set up the equation:

$c_1 (2t - 4t^2) + c_2 (6t^2 - 3t) = 0$

Expanding and combining terms:

$(2c_1 - 3c_2)t + (-4c_1 + 6c_2)t^2 = 0$

Equating each coefficient to zero:

1. $2c_1 - 3c_2 = 0$
2. $-4c_1 + 6c_2 = 0$

Dividing equation (2) by $-2$:

$2c_1 - 3c_2 = 0$

Since both equations are identical, this means there are infinitely many solutions for $c_1$ and $c_2$ that are not both zero. Therefore, these polynomials are linearly dependent. Answer: No.

Summary of Answers:

1. Yes (Linearly Independent)
2. Yes (Linearly Independent)
3. No (Linearly Dependent)

Would you like further details on any of these steps?

Related Questions:

1. What is the general method for determining linear independence in polynomial spaces?
2. How does the degree of the polynomial space affect linear independence?
3. Can we determine linear independence without setting up the coefficients directly?
4. How do the properties of polynomials relate to vector spaces?
5. What is the significance of having linearly independent polynomials in $\mathbb{P}_n$?

Tip:

When checking for linear independence in polynomials, setting up a system of equations based on coefficients is a systematic way to confirm dependency.