Prove by contradiction Z does not have a compositionseries. QZ does not have a composition series 3QZ has hon Maixmal sub Module 4Eve. Zz is not finitely cognreted. 5 Eevry vector space over askew find has basis. meday, serweltol pr not (3)

For those that you did not select as an element of P_2(R). Briefly explain why it was not an element of P_2(R).

23. Suppose R* = Span {v1, ..., V4}. Explain why {v1,..., V4} is a basis for R4.

th or of ne 17. Show that the set of all points in R2 lying on a line is a vector space with respect to the standard operations of vector ad- dition and scalar multiplication if and only if the line passes through the origin. 18. Show that the set of all points in R³ lying in a plane is a vector space with respect to the standard operations of vector addi- tion and scalar multiplication if and only if the plane passes. through the origin.

1.W.4 We'll work inside the vector space of polynomials in degree < 2, which is denoted P<2. Let P1 = 1, p2 = x + 2, and p3 = (x + 2)². Leť's think about Span{p1, P2, P3}. a) What polynomial is 3p1 + 2p2 – P3? b) I claim that a = a¡P1 + a2P2 + azp3 for some coefficients a1, a2, az in R. Find a1, a2, az. Hint: az = 0. c) I claim that a? = bịPi + b2p2 + b3p3 for some coefficients b1, b2, bz in R. Find b1, b2, bz. Hint: This problem isn't quite so cut and dry. Try to find three equations, one for each coefficient in the polynomial, and solve them for b1, b2, b3.

2.4 #8

Can any of the given sets of 3-vectors below span the 3-space? Why or why not? (a) [121] [231] [342]