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Thread: rank and null space

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    rank and null space

    just need to know how to solve a rank and null space of an 3x5 matrix. any help would be helpful thanks.
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    Quote Originally Posted by fatboy View Post
    just need to know how to solve a rank and null space of an 3x5 matrix. any help would be helpful thanks.
    The rank of a matrix is the dimension of its row or column space (both will be the same). Now, you could find a basis for the row space, and then the number of elements in the basis will be the rank of the matrix. To find a basis for the row space, you can use the fact that the nonzero rows of a matrix in row-echelon that is row-equivalent to another matrix A will actually form a basis for the row space of A.

    So, for example:

    $\displaystyle A = \left[\begin{array}{ccc}
    2 & -3 & 1\\5 & 10 & 6\\8 & -7 & 5
    \end{array}\right]$

    Put A in row echelon form to get:

    $\displaystyle B = \left[\begin{array}{ccc}
    1 & -\frac32 & \frac12\\ 0 & 1 & \frac15\\0 & 0 & 0
    \end{array}\right]$

    Now, since $\displaystyle B$ has two nonzero rows, you can conclude that $\displaystyle \text{rank}(A) = \text{dim}(\text{rowspace}(A)) = 2$.

    However, once you know the nullspace of a matrix, you usually don't need to go through the above process to find the rank.

    To find the nullspace of A, simply solve the system $\displaystyle A\textbf{x} = \textbf{0}$.

    For example, using the same matrix:

    $\displaystyle A = \left[\begin{array}{ccc}
    2 & -3 & 1\\5 & 10 & 6\\8 & -7 & 5
    \end{array}\right]$

    Augment this matrix with the 0 column vector and reduce:

    $\displaystyle \left[
    \begin{array}{ccc|c}
    2 & -3 & 1 & 0\\5 & 10 & 6 & 0\\8 & -7 & 5 & 0
    \end{array}
    \right]
    \rightarrow
    \left[
    \begin{array}{ccc|c}
    1 & 0 & \frac45 & 0\cr 0 & 1 & \frac15 & 0\cr 0 & 0 & 0 & 0
    \end{array}\right]
    $

    So,

    $\displaystyle
    \text{nullspace}(A) = N(A) =
    \left[\begin{array}{c}
    -\frac45t\\-\frac15t\\t
    \end{array}\right] =
    t\left[\begin{array}{c}
    -\frac45\\-\frac15\\1
    \end{array}\right]
    $

    The dimension of the nullspace is called the nullity. Knowing the nullity of a matrix allows you to find the rank very easily, without finding the row or column space: For any $\displaystyle m\times n$ matrix, $\displaystyle \text{rank}(A) + \text{nullity}(A) = n$.

    For our matrix, we see that $\displaystyle n = 3,\text{ and nullity}(A) = 1\Rightarrow\text{rank}(A) = 3 - 1 = 2$, which agrees with what we found earlier.

    I hope that helped!
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