# \ #
# "Proofs." #
# "(i) We can construct such a set of subspaces:" #
# "1)" \forall r \in RR, "let:" \qquad \quad V_r \ = \ \{ ( x, r x ) \in RR^2 | x \in RR \}. #
# "[Geometrically," \ V_r \ "is the line through origin of" \ RR^2, "of slope" \ r.] #
# "2) We will check that these subspaces justify assertion (i)." #
# "3) Clearly:" \qquad \qquad \qquad \qquad \qquad \qquad \qquad V_r sube RR^2. #
# "4) Check that:" \qquad \qquad V_r \ \ "is a proper subspace of" \ RR^2. #
# "Let:" \qquad u, v \in V_r, \ \alpha, \beta \in RR. \qquad \qquad \qquad \quad "Verify that:" \quad \alpha u+ \beta v \in V_r. #
# u, v \in V_r \ rArr \ u = ( x_1, r x_1 ), v = ( x_2, r x_2 ); \ "for some" \ x_1, x_2 \in RR #
# \qquad \qquad \qquad :. \qquad \quad \alpha u+ \beta v \ = \ \alpha ( x_1, r x_1 )+ \beta ( x_2, r x_2 ) #
# \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad = \ \alpha ( x_1, r x_1 )+ \beta ( x_2, r x_2 ) #
# \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad = \ ( \alpha x_1, \alpha r x_1 )+ ( \beta x_2, \beta r x_2 ) #
# \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad = \ ( \alpha x_1 + beta x_2, \alpha r x_1 + \beta r x_2 ) #
# \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad = \ ( \alpha x_1 + beta x_2, r (\alpha x_1 + \beta x_2 ) ) #
# \qquad \qquad \qquad \qquad \qquad \quad \quad = \ ( x_3, r x_3 ) \in V_r; \qquad "with" \ x_3 \ = \ \alpha x_1 + beta x_2. #
# "So:" \qquad \qquad \qquadu, v \in V_r, \ \alpha, \beta \in RR \quad rArr \quad \alpha u+ \beta v \in V_r. #
# "Thus:" \qquad \qquad \qquad \qquad \qquad \qquad \quad V_r \ "is a subspace of" \ RR^2. #
# "To see that" \ \ V_r \ \ "is non-zero, note that:" #
# \qquad \qquad \qquad \qquad \qquad \qquad (1, r ) in V_r, \ "and" \ (1, r ) \ne (0, 0).#
# "To see that" \ \ V_r \ \ "is proper," \ "note that" \ (1, r + 1 ) !in V_r: #
# (1, r + 1 ) in V_r rArr "(by construction of" \ V_r")" \quad r \cdot 1 = r + 1 #
# \qquad \qquad \qquad \qquad \qquad \qquad \ rArr r = r + 1, "plainly impossible." #
# "Thus:" \qquad \qquad \qquad V_r \ "is a non-zero, proper subspace of" \ RR^2. \qquad \qquad \qquad (1) #
# "5) Now show that there are infinitely-many such subspaces" \ V_r. #
# "Let:" \qquad \qquad r, s\in RR. \qquad \qquad \qquad \quad \ "We will show:" \qquad r \ne s \ rArr \ V_r \ne V_s. #
# "By definition:" \quad (1, r) = (1, r \cdot 1) \in V_r; \ \ (1, s) = (1, s \cdot 1) \in V_s. #
# "Clearly:" \qquad \qquad \qquad \qquad \qquad \ r \ne s \ rArr \ (1, r) \ne (1, s). #
# "Thus:" \qquad \qquad \qquad \qquad \qquad \qquad r \ne s \ \ rArr \ V_r \ne V_s. #
# "So each" \ \ r \in RR\ \ "produces a distinct subspace" \ V_r . #
# "This, together with (1), gives:" #
# "The family of subspaces:" \ \ \{ V_r | r \in RR \}, \ "is an infinite family" #
# "of non-zero, proper subspaces of" \ RR^2. \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad square #
# "(ii)This is actually easy. If the system is square, and the" #
# "coefficient matrix of the system in invertible, there will only be" #
# "the zero solution." #
# "Suppose:" \qquad \qquad \quad A \ "is a square, invertible matrix." #
# "Consider the homogeneous system:" #
# \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad A x = 0 . #
# "Thus, as" \ A \ "is invertible:" #
# \qquad \qquad \qquad \qquad \qquad \qquad \qquad \ A^{-1} \cdot A x = A^{-1} \cdot 0 . #
# \qquad \qquad \qquad \qquad :. \qquad \qquad \qquad \qquad \ \ \ I x = 0 . #
# \qquad \qquad \qquad \qquad :. \qquad \qquad \qquad \qquad \ \ \ \ \ x = 0 . #
# "Thus, the homogeneous system" \ \ A x = 0, \ "does not have a" #
# "non-zero solution." \qquad qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad square #
