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I don't understand why the rank = n - Rank-nullity theorem - nullity

  1. Feb 27, 2013 #1

    s3a

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    I don't understand why the rank = n -- Rank-nullity theorem -- nullity

    1. The problem statement, all variables and given/known data
    I'm working on #1 (the solutions are also included in that pdf) here ( http://ocw.mit.edu/courses/mathemat...-four-subspaces/exam-1/MIT18_06SCF11_ex1s.pdf ).

    2. Relevant equations
    Ax = b
    Ax = 0
    rank A+ nullity A = n

    3. The attempt at a solution
    For #1(a), I don't understand why the rank = n. I've been told to look at the Rank-nullity theorem which states that rank A+ nullity A = n but, I don't understand what nullity means exactly. In fact, I'm not very solid on the meaning of rank either (but, I kind of get it).

    Any input would be greatly appreciated!
     
  2. jcsd
  3. Feb 27, 2013 #2

    HallsofIvy

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    Re: I don't understand why the rank = n -- Rank-nullity theorem -- nul

    A linear transformation, A:U->V, maps vectors in a vector space, U, to a vector space V. It is fairly straight forward to show that the "range" of A, {v| v= Au for some u in U} is a subspace of V. The 'rank' of A is the dimension of that subspace. It is similarly straight forward to show that the null space of A, {u| AU= 0}. The 'nullity' of A is the dimension of the null space. One way of approaching the proof of the "rank-nullity" theorem is to choose a basis for the null space, [itex]\{u_1, u_2, ..., u_m\}[/itex] and expand it to a basis for U by adding vectors [itex]\{u_{m+1}, ..., u_n\}[/itex]. Show that every vector in the range of A can be written as a linear combination of [itex]\{Au_{m+1}, ..., Au_n\}[/itex] and so has dimension n- m.
     
  4. Feb 27, 2013 #3

    jbunniii

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    Re: I don't understand why the rank = n -- Rank-nullity theorem -- nul

    Two hints:

    The first statement,
    $$Ax = \left[ \begin{array}{c} 1 \\ 1 \\ 1 \end{array} \right] \textrm{ has no solutions}$$
    means that ##A## is not ...? (Injective, surjective?)

    The second statement,
    $$Ax = \left[ \begin{array}{c} 0 \\ 1 \\ 0 \end{array} \right] \textrm{ has exactly one solution}$$
    means that the kernel (null space) of ##A## is ...?
     
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