Final answer:
The vectors 1, x², and x²-2 in P₃ are linearly independent because the only solution to the equation that forms a linear combination of these vectors equal to the zero vector is the trivial solution where all scalar coefficients are zero.
Step-by-step explanation:
To determine whether the vectors 1, x², and x²-2 in the polynomial space P₃ are linearly independent, we can set up an equation where a linear combination of these vectors is equal to the zero vector. This equation looks like a * 1 + b * x² + c * (x²-2) = 0. Here, a, b, and c are scalar coefficients, and our task is to determine if the only solution for this equation is a = b = c = 0. Expanding the equation gives us a + bx² + cx² - 2c = 0.
To meet the condition of the zero vector (which is just 0 in the context of polynomials), the constant term and the coefficient of x² must both equal zero. Thus, we must solve the system of equations:
If this system has only the trivial solution a = b = c = 0, then the vectors are linearly independent. Through substitution, we can see that the only solution for this system is the trivial solution, thus, the vectors 1, x², and x²-2 are linearly independent in P₃.