Answer:
a) True
b) False
c) False
d) False
e) True
Step-by-step explanation:
a) Each basis of V has four vectors. Then any set of 5 vectors must be linear dependent (LD).
b) Suppose that [tex]\{v_1,v_2,v_3,v_4\}[/tex] is a basis of V. Considere the set [tex]A=\{v_1,\lambda_1v_1,\lambda_2v_1,v_2,v_3\}[/tex] where [tex]\lambda_1, \lambda_2[/tex] are scalars. The set has 5 vectors but [tex]V\neq span(A)[/tex] because [tex]v_4[/tex] is not belong to A and [tex]v_4[/tex] is linear independent of [tex]v_1[/tex]
c) Suppose that [tex]\{v_1,v_2,v_3,v_4\}[/tex] is a basis of V. Considere the set [tex]A=\{v_1,\lambda_1v_1,\lambda_2v_1,\lambda_3v_1\}[/tex] where [tex]\lambda_1, \lambda_2,\lambda_3[/tex] are scalars. A has four nonzero vectors but isn't a basis because is a LD set.
d) Suppose that [tex]\{v_1,v_2,v_3,v_4\}[/tex] is a basis of V. Considere the set [tex]A=\{v_1,\lambda_1v_1,\lambda_2v_1\}[/tex] where [tex]\lambda_1, \lambda_2[/tex] are scalars. A has 3 nonzero vectors but isn't a basis because is a LD set.
e) Since any basis of V must have 4 elements, then a set of three vectors cannot generate V.
