"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
