Results 1 to 13 of 13

Math Help - independant proof

  1. #1
    MHF Contributor
    Joined
    Nov 2008
    Posts
    1,401

    independant proof

    this is a two part question:
    v1..vn is a basis
    of R^n
    and there is a vector b which belongs to R^n and b differs the sero vector
    A.
    proof that {v1-vn,v2-vn,...,vn-1 - vn}

    by definition
    in order to prove that a group is independant
    we need to show the the only way for

    a1(v1-vn)+a2(v2-vn) ..+an(vn-1 -vn)=0

    is a1=..=an=0 all the coefficient hs to be zero
    so it rang a bell "i need to get a trivial solution "

    but trivilal solution is in Ax=0 system could be should if |A| differs zero.

    but A is a square matrices by difinition.
    how to construct from this single equation a square matrices?
    ??

    the second question:
    if v1..vn are solutions to Ax=b system then
    rho(A)=1
    ??
    rho(A) is the dimention of row or column space

    if v1..vn are solving this system
    then the dimention of the solution space is n dim(P(A))=n
    and from the formula where n=dim(P(A))+rho(A) we get n=n+rho(A)
    so i got
    rho(A)=0

    but i am asked to prove that rho(A)=1

    where is my mistake
    Last edited by transgalactic; June 22nd 2011 at 01:56 PM. Reason: fbdf
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Banned
    Joined
    Oct 2009
    Posts
    4,261

    re: independant proof

    Quote Originally Posted by transgalactic View Post
    this is a two part question:
    v1..vn is a basis
    of R^n
    and there is a vector b which belongs to R^n and b differs the sero vector
    A.
    proof that {v1-vn,v2-vn,...,vn-1 - vn}

    by definition
    in order to prove that a group is independant
    we need to show the the only way for

    a1(v1-vn)+a2(v2-vn) ..+an(vn-1 -vn)=0

    is a1=..=an=0 all the coefficient hs to be zero
    so it rang a bell "i need to get a trivial solution "

    but trivilal solution is in Ax=0 system could be should if |A| differs zero.

    but A is a square matrices by difinition.
    how to construct from this single equation a square matrices?
    ??

    the second question:
    if v1..vn are solutions to Ax=b system then
    rho(A)=1
    ??
    rho(A) is the dimention of row or column space

    if v1..vn are solving this system
    then the dimention of the solution space is n dim(P(A))=n
    and from the formula where n=dim(P(A))+rho(A) we get n=n+rho(A)
    so i got
    rho(A)=0

    but i am asked to prove that rho(A)=1

    where is my mistake


    Make yourself a favour and do use LaTeX for your questions (already more than 1,000!). It's extremely hard to read them as they are.

    Tonio
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Nov 2010
    Posts
    193

    re: independant proof

    Quote Originally Posted by tonio View Post
    Make yourself a favour and do use LaTeX for your questions (already more than 1,000!). It's extremely hard to read them as they are.

    Tonio
    In addition to spell check. Maybe grammar check as well, if such a thing is possible.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Joined
    Nov 2008
    Posts
    1,401

    re: independant proof

    at first i have what is given fro both question.

    then i write the firrst question

    and explain how i tried to solve it

    same thing for the second question
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member
    Joined
    Nov 2010
    Posts
    193

    re: independant proof

    Quote Originally Posted by transgalactic View Post
    at first i have what is given fro both question.

    then i write the firrst question

    and explain how i tried to solve it

    same thing for the second question
    Yes, that's all great, but that doesn't change the fact that your work is EXTREMELY difficult to read.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor
    Joined
    Nov 2008
    Posts
    1,401

    re: independant proof

    i will try to write in a more understandable way.

    i am given:
    v1..vn is a basis in R^n
    b\epsilon R^{n}
    b\neq\{0\}


    first question:
    proof that {v1-vn,v2-vn,...,vn-1 - vn} is an independant group

    how i tried to solve the first question:

    by definition:
    in order to prove that a group is independant,
    we need to show the the only way for the bottom equation to exist

    a1(v1-vn)+a2(v2-vn) ..+an(vn-1 -vn)=0

    is if a1=..=an=0 all the coefficient has to be zero.
    so it rang a bell "i need to get a trivial solution "

    but trivilal solution is in Ax=0 system could be should if |A| differs zero.

    but A is a square matrices by definition.
    how to construct from this single equation a square matrices?
    ??


    second question:

    if v1..vn are solutions to Ax=b system then
    \rho(A)=1
    ??

    how i tried to solve the second question:
    \rho(A) is the dimention of row or column space

    if v1..vn are solving this system
    then the dimention of the solution space is n, dim(P(A))=n
    and from the formula where n=dim(P(A))+ \rho(A) we get n=n+ \rho(A)
    so i got
    \rho(A)=0

    but i am asked to prove that \rho(A)=1

    where is my mistake
    ?
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Member
    Joined
    Nov 2010
    Posts
    193

    re: independant proof

    Well... for #2, the rank of the matrix certainly cannot be 0, unless it is the zero matrix. The problem is that the system is nonhomogeneous, so you can't directly apply that formula.

    Show first that v_1-v_2,v_2-v_3,v_1-v_3\in \mathrm{null}(A). Then show that any two of those form a basis for \mathrm{null}(A).

    For #1, you don't really need matrices. Just distribute everything out, collect like terms, and see what happens.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Banned
    Joined
    Oct 2009
    Posts
    4,261

    re: independant proof

    Quote Originally Posted by transgalactic View Post
    i will try to write in a more understandable way.

    i am given:
    v1..vn is a basis in R^n
    b\epsilon R^{n}
    b\neq\{0\}


    first question:
    proof that {v1-vn,v2-vn,...,vn-1 - vn} is an independant group

    how i tried to solve the first question:

    by definition:
    in order to prove that a group is independant,
    we need to show the the only way for the bottom equation to exist

    a1(v1-vn)+a2(v2-vn) ..+an(vn-1 -vn)=0

    is if a1=..=an=0 all the coefficient has to be zero.
    so it rang a bell "i need to get a trivial solution "

    but trivilal solution is in Ax=0 system could be should if |A| differs zero.

    but A is a square matrices by definition.
    how to construct from this single equation a square matrices?
    ??




    Several things:

    1) Stop calling "group" to sets of vectors. This is incorrect and confusing since groups are very different things.

    2) If all you have to do is to prove that \{v_1-v_n,\,v_2-v_n,\,...,\,v_{n-1}-v_n\} is an independent set , then do as follows:

    a_1(v_1-v_n)+a_2(v_2-v_n)+...+a_{n-1}(v_{n-1}-v_n)=0\Longrightarrow a_1v_1+a_2v_2+...+a_{n-1}v_{n-1}+(a_1+...+a_{n-1})v_0 . and now

    use that [tex]\[v_1,...,v_n\}[\tex] is an indep. set and deduce directly that a_1=a_2=...=a_{n-1}=0 .

    Tonio

    Pd. What's your mother tongue?


    second question:

    if v1..vn are solutions to Ax=b system then
    \rho(A)=1
    ??

    how i tried to solve the second question:
    \rho(A) is the dimention of row or column space

    if v1..vn are solving this system
    then the dimention of the solution space is n, dim(P(A))=n
    and from the formula where n=dim(P(A))+ \rho(A) we get n=n+ \rho(A)
    so i got
    \rho(A)=0

    but i am asked to prove that \rho(A)=1

    where is my mistake
    ?
    .
    Follow Math Help Forum on Facebook and Google+

  9. #9
    MHF Contributor
    Joined
    Nov 2008
    Posts
    1,401

    re: independant proof

    tonio:
    my mother tongue is russian but i am a student in israel so my course is in hebrew.
    thanks i got the idea

    topspin1617:
    so in the second question i cant use this formula
    n=dim(P(A))+ \rho(A)
    because P(A) represents the solution set of the homogeneus system.

    but here i got non homogeneus system.
    so what to do in the second question?
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Banned
    Joined
    Oct 2009
    Posts
    4,261

    re: independant proof

    Quote Originally Posted by transgalactic View Post
    tonio:
    my mother tongue is russian but i am a student in israel so my course is in hebrew.


    Ну, я понимаю, и я надеюсь, у вас есть удачи в учебе.

    באיזו אוניברסיטה אתה לומד? אני בוגר מהאוניברסיטה העברית בירושלים



    thanks i got the idea

    topspin1617:
    so in the second question i cant use this formula
    n=dim(P(A))+ \rho(A)
    because P(A) represents the solution set of the homogeneus system.

    but here i got non homogeneus system.
    so what to do in the second question?

    I'm answering in english for the sake of other participants: indeed, you can't use that

    formula because the set of all solutions to a non-homogeneous system is never a subspace.

    The second question is not that easy: if Ax=b\, ,\,b\neq 0 , then either \rho(A)=n and

    there's a unique solution, or else \rho(A)<n and there're more than one solution (in

    fact, infinite solutions if we're working over an infinite field).

    Now, what the question says, I think, is that the basis \{v_1,...,v_n\} is a solution to the non homogenoeus

    system, and from here we must deduce \rho(A)=1

    Now we must remember that any solution to a non-homogeneous system is of the form

    \alpha+\xi\, , \,\alpha=\mbox{ a particular solution to the non-homog. system and } \xi=\mbox { a solution of the asociated homogeneous sytem} ,

    so we can choose say v_n as a particular solution and then we get that

    \forall \,\,1\leq k <n\, , \,v_k=v_n+\xi_k \iff \{\xi_1,...,\xi_{n-1}\}=\{v_1-v_n,...,v_{n-1}-v_n\} , and this last set has

    dimension n - 1 by the first question, so...end the argument.

    Tonio
    Follow Math Help Forum on Facebook and Google+

  11. #11
    MHF Contributor
    Joined
    Nov 2008
    Posts
    1,401

    re: independant proof

    Ну, я понимаю, и я надеюсь, у вас есть удачи в учебе.

    באיזו אוניברסיטה אתה לומד? אני בוגר מהאוניברסיטה העברית בירושלים


    wow what are the odds for that to happen
    thanks i hope i get this succses in this course.

    i study at Ben gurion university
    Follow Math Help Forum on Facebook and Google+

  12. #12
    MHF Contributor
    Joined
    Nov 2008
    Posts
    1,401

    re: independant proof

    i though of a similar solution:
    v1..vn are solutions to the non homogeneus system Ax=b,
    if we subtract two solutions for the non homogeneus system then we get a solution to the homogeneus system AX=0.

    so from subtracting pairs from the n solutions to the non homogeneus we get n-1 solutions to the homogeneus system.

    so dim(P(A))=n-1
    now we apply the formula
    n=dim(P(A))+ \rho (A)
    n=n-1+ \rho (A)
    \rho (A)=1

    is this solution correct?
    Follow Math Help Forum on Facebook and Google+

  13. #13
    Member
    Joined
    Nov 2010
    Posts
    193

    re: independant proof

    Quote Originally Posted by transgalactic View Post
    i though of a similar solution:
    v1..vn are solutions to the non homogeneus system Ax=b,
    if we subtract two solutions for the non homogeneus system then we get a solution to the homogeneus system AX=0.

    so from subtracting pairs from the n solutions to the non homogeneus we get n-1 solutions to the homogeneus system.

    Okay up to here.

    so dim(P(A))=n-1
    now we apply the formula
    n=dim(P(A))+ \rho (A)
    n=n-1+ \rho (A)
    \rho (A)=1

    is this solution correct?
    So we have n-1 solutions to the system; i.e., n-1 elements of \mathrm{null}(A). This does NOT automatically force \mathrm{dim}\,\mathrm{null}(A)=n-1, however.

    tonio's solution works. I also have a solution, which looks different (though is probably the same idea anyway).

    We know that Av_1=\cdots =Av_n=b. As you have already noted, each of v_1-v_n,\ldots,v_{n-1}-v_n is an element of \mathrm{null}(A).

    We know these are linearly independent by the first part. Let's show that they also span \mathrm{null}(A).

    Let v\in \mathrm{null}(A) be arbitrary. Since v\in V and \{v_1,\ldots,v_n\} is a basis for V, we can write v=a_1v_1+\cdots +a_nv_n, for some a_1,\ldots,a_n\in \mathbb{R}. So we have

    0=Av=A(a_1v_1+\cdots a_nv_n)=a_1Av_1+\cdots +a_nAv_n=a_1b+\cdots +a_nb=(a_1+\cdots +a_n)b.

    Now, assume b\neq 0 (otherwise \mathrm{dim}\,\mathrm{null}(A)=n\Rightarrow A=0). This forces the scalar a_1+\cdots a_n=0 (the only way a scalar times a vector can be zero is if either the scalar or the vector is already zero). Solving for a_n, we have a_n=-(a_1+\cdots +a_{n-1}). Substituting this into the expression for v,

    v=a_1v_1+\cdots +a_{n-1}v_{n-1}-(a_1+\cdots +a_{n-1})v_n=a_1(v_1-v_n)+\cdots +a_{n-1}(v_{n-1}-v_n).

    Thus, v\in \mathrm{span}(v_1-v_n,\ldots,v_{n-1}-v_n), which is what we wanted to show.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. independant events
    Posted in the Advanced Statistics Forum
    Replies: 3
    Last Post: August 10th 2011, 07:06 PM
  2. independant proof
    Posted in the Advanced Algebra Forum
    Replies: 4
    Last Post: August 9th 2011, 02:36 PM
  3. Independant events
    Posted in the Statistics Forum
    Replies: 3
    Last Post: August 13th 2010, 01:36 PM
  4. independant events
    Posted in the Statistics Forum
    Replies: 3
    Last Post: April 22nd 2010, 03:33 AM
  5. Independant events.
    Posted in the Statistics Forum
    Replies: 1
    Last Post: October 24th 2009, 04:45 PM

Search Tags


/mathhelpforum @mathhelpforum