Question

A set V is given, together with definitions of addition and scalar multiplication. Determine which properties of a vector space are satisfied (Select all that apply.) V is the set of polynomials with real coefficients and degree 2 or less. Addition is defined by (a_2x^2 + a_1x + a_0) + (b_2x^2 + b_1x + b_0) = (a_0 + b_0)x^2 + (a_1 + b_1)x + (a_2 + b_2) and scalar multiplication by c(a_2x^2 + a_1 + a_0) = ca_0x^2 + ca_1x + ca_2. Property 1: If v_1 and v_2 are in V, then so is v_1 + v_2. Property 2: If c is a real scalar and v is in V, then so is cv. Property 3: There exists a zero vector 0 in V such that 0 + v = v for all v in V. Property 4: Property 3 holds and for each v in V there exists an additive inverse vector -v in V such that v + (-v) = 0 for all v in V. Property 5(a): For all v_1 and v_2 in V, we have v_1 + v_2 = V_2 + V_1. Property 5(b): For all v_1, v_2 and v_3 in V, we have (v_1 + v_2) + v_3 = v_1 + (v_2 + v_3). Property 5(c): For all v_1 and v_2 in V and real scalars c_1, we have c_1(v_1 + v_2) = c_1v_1 + c_1v_2. Property 5(d): For all v_1 in V and real scalars c_1 and c_2, we have (c_1 + c_2)v_1 = c_1v_1 + c_2v_1). Property 5(e): For all v_1 in V and real scalars c_1 and c_2, we have (c_1c_2)v_1 = c_1(c_2v_1). Property 5(f): For all v_1 in V, we have 1 - v_1 = v1. none of these

Answer 1

Answer:

Properties 1,2, 5(a) and 5(c) are satisfied, the rest of the properties arent valid.

Step-by-step explanation:

Note that both sum and scalar multiplication involves in exchanging the order from that main coefficient with the independent term before doing the standard sum/scalar multiplication.

Property 1 and 2 apply because by exchanging the order of 2 coefficients of a polynomial we obtain a polynomial of degree at most 2, and then we can conclude both properties are valid becuase standard sum of 2 polynomials of degree 2 or less or standard scalar multiplication of a polynomial with a real number will give as a result a polynomial of degree 2 or less.

Property 3 does not apply: Suppose that Property 3 is valid, lets call v = ax² +bx +c the neuter of V. Since v is the neuter, then 0 should be fixed by the neuted, therefore 0 = 0+v = (0x² + 0x + 0) + (ax² +bx +c) = cx²+b²+a.

0 is fixed by v only if c = b = a = 0. Thus, v = 0. If 0 is the neuter, then it should fix x², however 0 + x² = (0x²+0x+0) + (x²+0x+0) = 1. This is a contradiction because x² is not 1. We conclude that V doesnt have a neuter vector. This also means that property 4 doesn't apply either. A set without zero cant have additive inverse

Let v = v2x² + v1x + v0, w = w2x² + w1x + w0. We have that

• v + w = (v0+w0) * x² + (v1*w1) * x + (v2*w2)
• w + v = (w0+v0) * x² + (w1*v1) * x + (w2*v2)

Since the sum of real numbers is commutative, we conclude that v+w = w+v. Therefore, property 5(a) is valid.

Property 5(b) is not valid: we will introduce a counter example. lets use v = x², w = x²+1, z = 1, then

• (v+w)+z = (x²+2)+1 = 3x² + 1

• v + (w+z) = x² + (2x²+1) = x²+3

Since 3x²+1 ≠ x²+3, then the associativity rule doesnt hold.

Property 5(c) does apply. If v = v2x²+v1x+v0 and w = w2x²+w1x+w0, then we have that, for a real number c

• c*(v+w) = c*( (v0+w0)x² + (v1+w1)x + (v2+w2) ) = c*(v2+w2) x² + c*(v1+w1) x + c(v0+w0)
• c*v + c*w = (cv0x²+cv1x+cv2)*(cw0x²+cw1x+cw2) = (cv2+cw2)x²+(cv1+cw1)x+(cv0+cw0)

Note that both expressions are equal due to the distributive rule of real numbers. Also, you can notice that his property holds because in both cases we 'swap variables twice'. For this same argument neither of properties 5d and 5e apply, because on one term we swap variables just once and on the other term we swap variables twice. I will give an example with the vector x² + x and the scalars 1 and 2.

• (1+2)*(x²+x) = 3*(x² + x) = 3x + 3

• 1*(x²+x)+2*(x²+x) = (x+1)+(2x+2)  = 3x²+x (≠ 3x + 3)
• (1*2)*(x²+x) = 2*(x²+x) = 2x+2

• 1*(2*(x²+x)) = 1*(2x+2) = 2x²+2x (≠ 2x+2)

Property f doesnt apply due to the swap of variables. for example, if v = x², 1* v = 1*x² = 1 ≠ v.