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I don't understand the rank of a matrix!

  1. Sep 25, 2005 #1
    Hello everyone, can someone explain to me what the rank of a matrix is?
    I have the following:
    2 3 -2
    2 6 0
    -4 0 0
    Rank = 3;

    0 2 0 0
    0 0 0 -4
    0 0 0 0
    9 0 0 0
    rank = 3;


    1 2
    6 -3
    Rank = 2;

    I don't get it! any help would be great!
     
  2. jcsd
  3. Sep 25, 2005 #2
    Hey,

    The rank of a matrix is the number of linearly independent rows in the matrix.

    You can find the rank by performing Gaussian elimination. The rank will then be the number of non-zero rows in the resulting matrix.
     
  4. Sep 25, 2005 #3
    To compute the rank of a matrix, do the following.

    1) Let [tex] A [/tex] be a [tex] m\times n[/tex] matrix.
    2) Perform gaussian-elimination on [tex] A [/tex]
    3) Count the number of non zero columns. This number is the rank of the matrix.

    We'll start with your first example.
    1) [tex] A = \left[ \begin{array}{ccc}2 & 3 & -2 \\2 & 6 & 0 \\-4 & 0 & 0 \\\end{array}\right] [/tex]

    2) Perform the gaussian-elimination on [tex] A [/tex]
    [tex] A' = \left[ \begin{array}{ccc}1 & 0 & 0 \\0 & 1 & 0 \\0 & 0 & 1 \\\end{array}\right] [/tex]

    3) Now we count the number of non-zero columns:
    [tex] A' = \left[ \begin{array}{ccc}(1) & 0 & 0 \\0 & (1) & 0 \\0 & 0 & (1) \\\end{array}\right] [/tex]

    Rank = 3

    ----
    Now, what does the rank mean? Well the rank of a matrix lets you know the number of columns that cannot be written as a linear combination of each other.

    If you consider [tex] A' [/tex] to be a matrix of vectors, we would have the following column vectors.

    [tex] \vec{V_1} = (1,0,0) [/tex]
    [tex] \vec{V_2} = (0,1,0) [/tex]
    [tex] \vec{V_3} = (0,0,1) [/tex]

    These vectors are ALL linearly independent of each other. Which means that the column vectors in [tex] A [/tex] are all linearly independent of each other.
     
  5. Sep 25, 2005 #4
    Just don't forget that the rank only tells you the number of independent columns. If you are trying to find the spanning set, you must go back to the ORIGINAL matrix.
     
  6. Sep 25, 2005 #5
    Awesome! thanks alot guys! good explanation!! :biggrin:
     
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