# "This is a good question -- answer is worth keeping handy." #
# "Fortunately, the proof is very simple. We will create a" #
# "homomorphism of the additive groups, and then apply the" #
# "Fundamental Homomorphism Theorem." #
# "First, a caution. In a quotient of any algebraic systems, the" #
# "denominator set is, of course, a subset of the numerator set." #
# "However, what is asked to be shown, refers to the quotient" #
# { RR^n } / { RR^m }. \ \ "The vectors in" \ RR^n \ "have length" \ n, "while the vectors in" \ RR^m \ "have length" \ m. \ \ "As these are different lengths, the" #
# "denominator," \ RR^m, \ "cannot be a subset of the numerator," \ RR^n. #
# "So we must correct the statement to be shown." #
# "(Note that the case where" \ n = m, "in which the lengths of the" #
# "vectors of the two sets are the same, will not need to be" #
# "handled separately; the correction we will make, in what is" #
# "to be shown, will include this case, automatically.)" #
# "Here is how to make the corrected statement." #
# "Let:" \qquad \hat{ RR^m } \ = \ "the subset of vectors of" \ \ RR^n \ "defined by:" #
# \hat{ RR^m } \ = #
# \{ ( \overbrace{ 0, ... , 0 }^{ n - m }, \ \overbrace{ a_{ n - m + 1 }, ..., \ a_n }^{ m } ) | a_{ n - m + 1 }, \ ... , \ a_n \in RR \}. \quad \ (I) #
# "We can think of the vectors in" \ \ hat{ RR^m } \ \ "as the vectors of" \ \ RR^m #
# "with" \ ( n - m ) \quad \quad 0 "'s inserted in the front. So they are" #
# "essentially the same algebraic system. Precisely, we clearly" #
# "have:" \qquad \hat{ RR^m } \ ~~ \ RR^m \ \ "[exercise], by using the map:" #
# \qquad ( hat{ RR^m }, + ) \ rarr \ ( RR^m, + ); #
# \qquad \qquad \quad \ ( \overbrace{ 0, ... , 0 }^{ n - m }, \ \overbrace{ a_{ n - m + 1 }, ..., \ a_n }^{ m } ) \mapsto ( \overbrace{ a_{ n - m + 1 }, ..., \ a_n }^{ m } ). #
# "With the correct subset of" \ \ RR^n \ \ "to use, now defined, we will" #
# "show the corrected statement:" #
# \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad RR^n / hat{ RR^m } \quad ~~ \quad RR^{ n - m }. \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad (II) #
# "Ok, with this caution addressed, let's go back to beginning," #
# "and start over. We are given that:" \ \ m, n \in NN, \quad "and" \quad m <= n. #
# "So, in particular, we have that:" \qquad n - m \in NN. #
# "Consider the map:" #
# \quad \ \pi: \ \ ( RR^n, + ) \ rarr \ ( RR^{ n - m }, + ) #
# \qquad \qquad \qquad \qquad \pi: ( \overbrace{ a_1, \ a_2, \ ..., a_{ n - m } }^{ n - m }, \ \overbrace{ a_{ n - m + 1 }, ..., \ a_n }^{ m } ) #
# \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \mapsto ( a_1, \ a_2, \ ..., \ a_{n-m} ). #
# "We can think of this map as taking a vector in" \ \ RR^n, "and" #
# "deleting its last" \ m \ "entries. This map is sometimes" #
# "called A Projection of" \ \ RR^n \ "onto" \ \ RR^{ n - m }. #
# "We can visualize it like this:" #
# \qquad \qquad \qquad \pi: ( \overbrace{ a_1, \ a_2, \ ..., a_{ n - m } }^{ n - m }, \ \overbrace{ a_{ n - m + 1 }, ..., \ a_n }^{ "delete last" \ m \ "entries" } ) \qquad \qquad \qquad \qquad \quad (I) #
# \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \mapsto ( a_1, \ a_2, \ ..., \ a_{n-m} ). #
# "Although the work we will do below is relatively" #
# "straightforward, the long vectors may make it appear" #
# "complicated. Remembering the visual above in (I), will make" #
# "things much, much easier !! Whenever we are working with," #
# "or calculating, a quantity like" \quad \pi( vec(v) ) \ , "it will be very useful" #
# "to remember this visual." #
# "Let me reiterate that the work here is actually simple and" #
# "straightforward. The long vectors may make it appear " #
# "complicated -- it is not." #
# "We will show the map," \ \pi, \ "is a homomorphism of the" #
# "additive groups," \quad ( RR^n, + ) \quad "and" \quad ( RR^{ n - m }, + ). \ \ "Then we will" #
# "calculate its kernel, and apply the Fundamental" #
# "Homomorphism Theorem." #
# "1) We show now " \ \pi \ "is a homomorphism of the additive" #
# \qquad \quad \ "groups." #
# "Let:" \qquad vec{a} \ = \ ( \overbrace{ a_1, \ a_2, \ ... , a_{n-m} }^{ n - m }, \ \overbrace{ \ a_{n-m +1}, ... , \ a_n }^{ m } ), \qquad "and" #
# \qquad \qquad \qquad vec{b} \ = \ ( \overbrace{ b_1, \ b_2, \ ... , \ b_{n-m} }^{ n - m }, \overbrace{ b_{n-m +1}, \ ... , \ b_n}^{ m } ). #
# "We compute:" \qquad \pi( vec{a} - vec{b} ). #
# \pi( vec{a} - vec{b} ) \ = \ \pi[ \ ( \overbrace{ a_1, \ a_2, \ ..., a_{n-m} }^{ n - m }, \ \overbrace{ a_{n-m +1}, \ ..., \ a_n }^{ m } ) #
# \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad \ - ( \overbrace{ b_1, \ b_2, \ ..., \ b_{n-m} }^{ n - m }, \ overbrace{ b_{n-m +1}, \ ..., \ b_n }^{ m } ) \ ] #
# = \ \pi[ \ ( \overbrace{ a_1 - b_1 }, \ \overbrace{ a_2 - b_2 }, \ ... , \ \overbrace{ a_{n-m} -b_{n-m} }, #
# \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \ \ \ \overbrace{ a_{n-m +1} - b_{n-m +1} }, \ ..., \ \overbrace{ a_n - b_n } ) \ ] #
# \qquad \qquad " ... delete last" \quad m \quad "entries in previous ... "#
# = \ ( \overbrace{ a_1 - b_1 }, \ \overbrace{ a_2 - b_2 }, \ ..., \ \overbrace{ a_{n-m} -b_{n-m} } ) #
# = \ ( a_1, a_2, ..., a_{n-m} ) - ( b_1, b_2, ..., b_{n-m} ) #
# \qquad "continuing, and using the definition, in reverse," #
# \qquad \qquad \quad "of the map" \ \ \pi ":" #
# = \ \pi [ \ ( \overbrace{ a_1, \ a_2, \ ..., a_{n-m} }^{ n - m }, \ \overbrace{ a_{n-m +1}, \ ..., \ a_n }^{ m } ) \ ] #
# \qquad \qquad \qquad - \pi [ \ ( \overbrace{ b_1, \ b_2, \ ..., \ b_{n-m} }^{ n - m }, \ overbrace{ b_{n-m +1}, \ ..., \ b_n }^{ m } ) \ ] #
# = \pi( vec{a} ) - \pi( vec{b} ). #
# "So, from top to bottom here, we have shown:"#
# \qquad \qquad \qquad \qquad \qquad \quad \pi( vec{a} - vec{b} ) \ = \ \pi( vec{a} ) - \pi( vec{b} ). #
# "Thus:" \quad \pi \quad "is a homorphism of the additive groups:" #
# \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad ( RR^n, + ) , \ ( RR^{ n - m }, + ). \qquad \qquad \qquad \qquad \qquad \qquad \quad \(II) #
# "2) So, we have immediately, by the Fundamental" #
# \qquad \qquad "Homomorphism Theorem:" #
# \qquad \qquad \qquad \qquad \qquad \qquad \qquad RR^n / { ker(\pi) } \quad ~~ \quad Im( \pi ). \qquad \qquad\qquad \qquad \qquad \qquad \quad \quad \ \ \ (III) #
# "We will show that:" \qquad ker(\pi) \ = \ hat{RR^m} \quad "and" \quad Im( \pi ) \ = \ RR^{ n - m }; #
# "which will establish the desired result." #
# "a) Let:" \qquad \qquad vec{ z } \in RR^n \qquad "and" \qquad vec{ z } \in ker( \pi ). #
# \qquad o. \qquad vec{ z } \in RR^n \quad hArr #
# \qquad \qquad \qquad \qquad \quad " vec{ z } \ = \ ( \overbrace{ z_1, \ z_2, \ ..., z_{n-m} }^{ n - m }, \overbrace{ \ z_{n-m +1}, ..., \ z_n }^{ m } ), #
# \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad "for some" \quad z_1, \ z_2, \ ..., \ z_n \in RR. #
# \qquad o. \qquad vec{ z } \in ker( \pi ) \quad hArr #
# \ \pi [ ( ( \overbrace{ z_1, \ z_2, \ ..., z_{n-m} }^{ n - m } , \overbrace{ \ z_{n-m +1}, ... , \ z_n }^{ m } ) ) ] \ = \ ( \overbrace{ 0, 0, ... , 0 }^{ n - m } ). #
# \qquad \qquad "continuing, and using the definition of the map" \ \ \pi \ - #
# \qquad \qquad \qquad \qquad "i.e., delete the last" \ m \ "entries, we have:" #
# \qquad \qquad \qquad \qquad \ ( \overbrace{ z_1, \ z_2, \ ..., z_{n-m} }^{ n - m } ) \ = \ ( \overbrace{ 0, 0, ... , 0 }^{ n - m } ). #
# "Hence, we have:" #
# \qquad \qquad \qquad \qquad \qquad \qquad \quad \ z_1 = 0, \ z_2 = 0, \ ... , \ z_{n-m} = 0. #
# "Now let's put this information back into" \ \ vec{ z }:" #
# \qquad \qquad \qquad \qquad \quad " vec{ z } \ = \ ( \overbrace{ z_1, \ z_2, \ ..., z_{n-m} }^{ n - m }, \overbrace{ \ z_{n-m +1}, ..., \ z_n }^{ m } ) #
# \qquad \qquad \qquad \qquad \quad " vec{ z } \ = \ ( \overbrace{ 0, 0, ... , 0 }^{ n - m }, \overbrace{ \ z_{n-m +1}, ..., \ z_n }^{ m } ). #
# "Now, recalling the definition of the set:" \quad \hat{ RR^m }, \ "in (I) above," #
# "we have:" #
# \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \ vec{ z } \in \hat{ RR^m }. #
# "Hence, from top to bottom in this part, we have:" #
# \qquad \qquad \qquad \qquad \qquad \qquad \qquad vec{ z } \in ker( \pi ) \quad hArr \quad vec{ z } \in \hat{ RR^m }. #
# "And so, we have:" #
# \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad ker( \pi ) \ = \ \hat{ RR^m }. #
# "And now that we have found" \ \ ker( \pi ), "we substitute it back" #
# "into the result of the Fundamental Homomorphism" #
# "Theorem we had in (III) here. We get:" #
# \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quadRR^n / \hat{ RR^m } \quad ~~ \quad Im( \pi ). \qquad \qquad\qquad \qquad \qquad \qquad \qquad (IV) #
# "This is now much closer to the desired result. We are almost" #
# "done. We will now find" \ Im( \pi ). \ \ "This will be easy." #
# "b) Let:" \qquad \qquad vec{ t } \in RR^{ n- m }. #
# "So then:" #
# \qquad vec{ t } \ = \ ( \overbrace{ t_1, \ t_2, \ ..., t_{n-m} }^{ n - m } ), \qquad "for some" \quad t_1, \ t_2, \ ..., \ t_n \in RR. #
# "Now let:" \qquad \qquad vec{ T } \in RR^n, \quad"where we define" \ vec{ T } \ "as follows:" #
# \qquad \qquad \qquad\qquad \qquad vec{ T } \ = \ ( \overbrace{ t_1, \ t_2, \ ..., t_{n-m} }^{ n - m }, \overbrace{ \ 1, ..., \ 1 }^{ m } ) . #
# "So we have:" #
# \qquad o. \qquad vec{ T } \in RR^n; #
# \qquad o. \qquad \pi (vec{ T } ) \ = \ \pi [ ( ( \overbrace{ t_1, \ t_2, \ ..., t_{n-m} }^{ n - m }, \overbrace{ \ 1, ..., \ 1 }^{ m } ) ) ] #
# \qquad \qquad \qquad \qquad \qquad \qquad \ \ = \ \pi [ ( ( \overbrace{ t_1, \ t_2, \ ..., t_{n-m} }^{ n - m }, \overbrace{ \ 1, ..., \ 1 }^{ "delete" } ) ) ] #
# \qquad \qquad \qquad \qquad \qquad \qquad \ \ = \ ( \overbrace{ t_1, \ t_2, \ ..., t_{n-m} }^{ n - m } ) #
# \qquad \qquad \qquad \qquad \qquad \qquad \ \ = \ vec{ t }, \qquad \qquad \qquad \ \ "by definition of" \ vec{ t } \ "above here." #
# "So from the above, we have:" #
# \qquad \qquad \qquad \qquad \qquad \qquad \quad vec{ t } \ = \ \pi (vec{ T } ) qquad "and" \qquad vec{ T } \in RR^n. #
# "Thus, by definition of the image of a map:" #
# \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad vec{ t } \ in Im( \pi ). #
# "As" \ vec{ t } \ "was taken arbitrarily in" \ RR^{ n- m }, "we have:" #
# \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad \ \ \ RR^{ n- m } sube Im( \pi ). #
# "But since" \ \ pi \ \ "maps into" \ \ RR^{ n- m } \ , \ "by definition of the image of a" #
# "map, we have:" #
# \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad \ \ \ Im( \pi ) sube RR^{ n- m }. #
# "So we have now:" #
# \qquad \qquad \qquad \quad \quad \ Im( \pi ) sube RR^{ n- m } \qquad "and" \qquad RR^{ n- m } sube Im( \pi ). #
# "Thus:" #
# \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad \ \ \ Im( \pi ) \ = \ RR^{ n- m }. #
# "Now that we know" \ Im( \pi ), "we may substitute this back into" #
# "our intermediate, and major, result in (IV). We get:" #
# \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad RR^n / \hat{ RR^m } \quad ~~ \quad RR^{ n- m }. #
# "This is our desired result !!" \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad square #
