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Math Help - reconstruct matrix from its solution?

  1. #1
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    reconstruct matrix from its solution?

    I am stuck on this question, I guess my brain only mastered one way matrix manipulation. I would love to get to the bottom of the things here, appreciate your help! This is an exam practice question, and I don't have a solution to it.


    Question.

    A system of linear equations Ax=d is known to have the following solution:

    x=(1,2,0,-1,0)^T+s(2,1,1,0,0)^T+t(1,1,0,-1,1)^T.

    Assume that A is an m x n matrix. Let c_1, c_2, ... c_n denote the columns of A.

    Answer the following questions, or, if there is insufficient information to answer the question, say so.

    (1) What number is n?
    (2) What number is m?
    (3) What (number) is the rank of A?
    (4) Write down a basis for the nullspace of A, N(A).
    (5) Which columns c_i form a basis of the range, R(A)?
    (6) Write down an expression for d as a linear combinations of columns c_i.
    (7) Write down a non-trivial linear combination of columns c_i which is equal to the zero vector.


    My thoughts:

    Ax=d means A(mx5 matrix) x x(5x1 vector) = d(mx1 vector)

    Further, if I put the given information into equation format, I get:

    x_1=1+2s+t
    x_2=2+s+t
    x_3=s
    x_4=-1-t
    x_5=t

    I have two free variables - x_3 and x_5 (s and t respectively) - therefore there are two zero rows in A.

    Also, the columns 1, 2 and 4 are columns containing pivot variables x_1, x_2,x_4. Does it mean that there are 3 rows containing pivot variables also???

    Finally, is there a way to write down A from the given solution? If I could do that, I'd answer all the questions below in an instant...


    So far:

    (1) n=5

    (2) ??? 5 or 3, and I am not sure how to prove either

    (3) ???

    (4) columns containing free variables would form the basis for nullspace: c_3, c_5

    (5) columns containing pivot variables would the basis for the range, ie c_1, c_2, c_4

    (6) d as linear combination of the columns [tex]c_i[\math] would be x_1c_1+x_2c_2+...+x_5c_5 except that I probably need to insert the values for c's (((

    (7) to find c's for Ax=0, I think I would need to solve Ax=0 for x (when I know A).
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  2. #2
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    For the ones you said "???", i would suggest something like this (but im not a maths student so i dont know how much detail is supposed to be in the answer):

    (2) i dont think you can deduce the number of rows in A. Proof by contradiction: Suppose you knew A and it had m rows, if you added a copy of the last row the same vector x would still solve Ax=d. Alternatively (and trivially) add a row of zeros to both A and d. All that can be deduced is the number of linearly independant rows in A (which is the next question)

    (3) 3

    There is only one solution iff there are 5 unique rows (Rouché–Capelli theoum according to wikipedia). Your system has infinitely many solutions (corresponding to all possible values of s and t). each "spare" variable means the rank of A must be reduced by 1.

    This could be proved by adding 2 new (linearly independant) equations for s,t and noting that there is then only 1 solution.


    I'll leave it to someone better than me to verify your other answers.
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  3. #3
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    Thank you! I felt that there was somehow a connection between (2) and (3) - total number of rows and rank - but I can see clearly now why we cannot know the total number of rows, only number of linearly independent rows. I am half way there!!

    PS that always puzzled me, by the way, what's the point of having zero rows in a matrix? or having a matrix with linearly dependent rows, in other words. they don't add any information, apart from dimension of the column space may be, but it's zero anyway in that 'additional' dimension.
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  4. #4
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    duplicate rows are informative in some practical situations, (statistics comes to mind, but i expect there are others). For example in OLS estimation adding a duplicate row will move the estimates towards values consistent with that row.
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