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Linear Algebra - Column and Null Space - Take 2

  1. Oct 26, 2009 #1
    1. The problem statement, all variables and given/known data

    Ax=b where,

    A = 1 -1
    .....-1 1

    2. Relevant equations

    a) Find Null Space N(A) and Column Space C(A)
    b) For which vectors b does the system Kx=b have a solution?
    c) How many solution x does the system have for any given b

    3. The attempt at a solution

    a)

    For Null Space, I got x = 1
    ..................................1
    For Column Space, I got b = span ( 1 -1 )
    ................................................-1 1

    b)

    This is once again, equal to the column space.

    c)

    Following the similar approach as before,

    Let Ax = b and Ay = b be two equations.

    Ax - Ay = b - b
    A(x - y) = 0

    Since we solved Ax=0 for Null Space, x = 1
    ........................................................1

    So, x - y = 1
    ................1

    Now, I know it isn't only equal to that vector. Correct? Null Space is all scalar multiple and sums of that vector.

    So can I now say that x - y = c 1 ?
    ............................................1

    Also by looking at that, I want to say that there are an infinite number of solutions for a given b. Because, there is an infinite number of ways to combine x and y to get the vector one one.

    I am confident this time. Can someone concur?

    Also, is there a thread on this forum somewhere that teaches you how to post matrices and other mathematical equations so to say the right way? Maybe a forum where you can just post a junk thread to practice?

    Thanks again, I love the input! :)
     
  2. jcsd
  3. Oct 26, 2009 #2

    Dick

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    Science Advisor
    Homework Helper

    At this point it is getting hard to read because of the eccentric formatting. I think there is a thread on how to do TeX formatting. Once again, I can't find it. If anyone knows, please help me out here. I'll give you two suggestions. i) Do as I do. Write a 2x2 matrix as [[a,b],[c,d]], where [a,b] is the first row and [c,d] is the second row. Write a vector as [a,b], if you really have to emphasize it's column vector write [a,b]^T. That's the lazy solution, but it will at least save you a lot of ........... stuff. ii) find a post where someone has formatted a matrix and right click on the matrix. It should bring up a box showing you the tex code they used to do it. That's the smart way to do it. Don't be afraid to experiment in your own post. I'll try and find the tex lesson again.
     
  4. Oct 27, 2009 #3

    Dick

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  5. Oct 28, 2009 #4
    Bringing this up with a reconstruction of the original post. Format updated with LaTex for a clearer presentation. Thanks Dick!

    1. The problem statement, all variables and given/known data


    [tex]

    \begin{gather*}
    Ax = b\\
    \end{gather*}

    [/tex]

    Where,

    [tex]

    A = \left[
    \begin{array}{cc}
    1 & -1 \\
    -1 & 1
    \end{array}
    \right]

    [/tex]

    a) Find Null Space N(A) and Column Space C(A)
    b) For which vectors b does the system Kx=b have a solution?
    c) How many solution x does the system have for any given b

    2. The attempt at a solution

    a)

    For Null Space, I got

    [tex]

    x = \left[
    \begin{array}{cc}
    1\\
    1
    \end{array}
    \right]

    [/tex]

    For Column Space, I got

    [tex]

    b = span \left( \left[
    \begin{array}{cc}
    1 \\
    -1
    \end{array}
    \right] ,
    \left[
    \begin{array}{cc}
    -1\\
    1
    \end{array}
    \right]
    \right)

    [/tex]
    b)

    This is once again, equal to the column space.

    c)

    Following the similar approach as before,

    Let [tex]

    \begin{gather*}
    Ax' = b\\
    \end{gather*}

    [/tex] and [tex]

    \begin{gather*}
    Ay = b\\
    \end{gather*}

    [/tex] be two equations.

    Then,

    [tex]

    \begin{gather*}
    Ax'- Ay = b-b\\
    \end{gather*}

    [/tex]

    [tex]

    \begin{gather*}
    A(x'-y) = 0\\
    \end{gather*}

    [/tex]

    Since we solved [tex]

    \begin{gather*}
    Ax = 0\\
    \end{gather*}

    [/tex] for Null Space,

    [tex]

    x = \left[
    \begin{array}{cc}
    1\\
    1
    \end{array}
    \right]

    [/tex]

    [tex]

    x'-y = \left[
    \begin{array}{cc}
    1\\
    1
    \end{array}
    \right]

    [/tex]

    Now, I know it isn't only equal to that vector. Correct? Null Space is all scalar multiple and sums of that vector.

    So can I now say that

    [tex]

    x'-y = \left[
    \begin{array}{cc}
    1\\
    1
    \end{array}
    \right]
    [/tex]

    Also by looking at that, I want to say that there are an infinite number of solutions for a given b. Because, there is an infinite number of ways to combine x and y to get the vector one one.

    I am confident this time. Can someone concur?

    Thanks again, I love the input! :)
     
    Last edited: Oct 28, 2009
  6. Oct 28, 2009 #5

    Dick

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    Science Advisor
    Homework Helper

    Ok, forgive the less pretty presentation here. The null space is span([1,1]), right? Not just [1,1]. I think you probably know this, just checking. Yes, the column space is span([1,-1],[-1,1]), but that's the same as span([1,-1]), right? [1,-1] and [-1,1] aren't linearly independent. Now if Ax=b and Ay=b, then, yes, A(x-y)=0. So (x-y) is in the null space. But that doesn't mean it's [1,1], it means it's c*[1,1] for some constant c, as you said in the phrase "Null Space is all scalar multiple and sums of that vector."
    So, sure, there are an infinite number of solutions. If x is a solution of Ax=b then so is x+c*[1,1] for any constant c. Some of what you've actually written doesn't quite reflect what I think you know.
     
  7. Oct 28, 2009 #6
    I did not "know" it was span for null space, but I did know it was all multiples and sums of [1,1]. Bit redundant huh? Makes complete sense though. I also knew it was multiplied by a scalar c, as I posted in original post. Just did not put it in there when I the fancy print, oops.

    All is good. I submitted the h/w last night, and I did put infinite number of solutions for that reason. Just wanted to see if I did it right. Thank you! :)
     
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