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Math Help - Kernel of Transformation?

  1. #1
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    Kernel of Transformation?

    I have the matrix
    1st line [1 0 1 -2]
    2nd line [1 1 -3 4]
    3rd line [3 1 -1 0]
    4th line [4 2 -4 4]
    5th line [0 1 -4 6]

    I need to transform it from R5 to R4, ann then find the base nullity. How do I go about doing this and if possible show a working out. Cheers
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  2. #2
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    Quote Originally Posted by LooNiE View Post
    I have the matrix
    1st line [1 0 1 -2]
    2nd line [1 1 -3 4]
    3rd line [3 1 -1 0]
    4th line [4 2 -4 4]
    5th line [0 1 -4 6]

    I need to transform it from R5 to R4, ann then find the base nullity. How do I go about doing this and if possible show a working out. Cheers
    Let M be that matrix if you define T: \mathbb{R}_4 \to \mathbb{R}_5 by T(\bold{x}) = M\bold{x} you get a linear transformation.

    The nullity of T is \text{dim} \left( \{ \bold{x}\in \mathbb{R}_4 |T(\bold{x}) = \bold{0} \} \right).

    The first step is to turn the system M\bold{x} = \bold{0} to row-reduced echelon form.

    Can you do that first?
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  3. #3
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    Quote Originally Posted by ThePerfectHacker View Post
    Let M be that matrix if you define T: \mathbb{R}_4 \to \mathbb{R}_5 by T(\bold{x}) = M\bold{x} you get a linear transformation.

    The nullity of T is \text{dim} \left( \{ \bold{x}\in \mathbb{R}_4 |T(\bold{x}) = \bold{0} \} \right).

    The first step is to turn the system M\bold{x} = \bold{0} to row-reduced echelon form.

    Can you do that first?
    Ok i tried reducing it, and this is what i got:
    [1 0 1 -2]
    [0 1 -4 -2]
    [0 0 0 8]
    [0 0 0 16]
    [0 0 0 8]
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  4. #4
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    Quote Originally Posted by LooNiE View Post
    Ok i tried reducing it, and this is what i got:
    [1 0 1 -2]
    [0 1 -4 -2]
    [0 0 0 8]
    [0 0 0 16]
    [0 0 0 8]
    When you have the system M\bold{x} = \bold{0} and you are solving for \bold{x}.
    The matrix you end up with is actually:
    \begin{bmatrix} 1&0&1&-2&0 \\ 1&1&-3&4&0 \\ 3&1&-1&0&0 \\ 4&2&-4&4& 0 \\ 0&1&-4&6&0 \end{bmatrix}

    Now if you done the row-reductions (assuming you were correct!) we get:
    \begin{bmatrix} 1&0&1&-2&0\\0&1&-4&-2&0\\0&0&0&8&0\\0&0&0&16&0\\0&0&0&8&0 \end{bmatrix}
    You can bring the bottom three rows to:
    \begin{bmatrix} 1&0&1&-2&0\\0&1&-4&-2&0\\0&0&0&1&0\\0&0&0&1&0\\0&0&0&1&0 \end{bmatrix}

    However this matrix is still now in row-reduced echelon form.
    We will add the (3rd row \times 2) to 2nd row and 1st row.
    \begin{bmatrix} 1&0&1&0&0\\0&1&-4&0&0\\0&0&0&1&0\\0&0&0&1&0\\0&0&0&1&0 \end{bmatrix}

    Finally, add (2nd row \times -\tfrac{1}{4}) to 1st row:
    \begin{bmatrix} 1&0&0&0&0\\0&1&-4&0&0\\0&0&0&1&0\\0&0&0&1&0\\0&0&0&1&0 \end{bmatrix}

    This tells you that:
    x_1 = 0
    x_2 - 4x_3 = 0
    x_4 = 1.

    Thus, x_1=0,x_4=1,x_2 = 4x_3 (notice that x_3 is "free").
    This means if t\in \mathbb{R} then \bold{x} = \begin{bmatrix}0\\4t\\t\\1 \end{bmatrix} solves M\bold{x} = \bold{0}, and conversely.

    Thus, the nullspace is \left\{ \begin{bmatrix}0\\4t\\t\\1 \end{bmatrix} : t\in \mathbb{R} \right\} = \left\{ t \begin{bmatrix}0\\4\\1\\1 \end{bmatrix} : t\in \mathbb{R}\right\}

    We see the dimension is one, and so the nullity is one.
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  5. #5
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    Exclamation

    Quote Originally Posted by ThePerfectHacker View Post
    When you have the system M\bold{x} = \bold{0} and you are solving for \bold{x}.
    The matrix you end up with is actually:
    \begin{bmatrix} 1&0&1&-2&0 \\ 1&1&-3&4&0 \\ 3&1&-1&0&0 \\ 4&2&-4&4& 0 \\ 0&1&-4&6&0 \end{bmatrix}

    Now if you done the row-reductions (assuming you were correct!) we get:
    \begin{bmatrix} 1&0&1&-2&0\\0&1&-4&-2&0\\0&0&0&8&0\\0&0&0&16&0\\0&0&0&8&0 \end{bmatrix}
    You can bring the bottom three rows to:
    \begin{bmatrix} 1&0&1&-2&0\\0&1&-4&-2&0\\0&0&0&1&0\\0&0&0&1&0\\0&0&0&1&0 \end{bmatrix}

    However this matrix is still now in row-reduced echelon form.
    We will add the (3rd row \times 2) to 2nd row and 1st row.
    \begin{bmatrix} 1&0&1&0&0\\0&1&-4&0&0\\0&0&0&1&0\\0&0&0&1&0\\0&0&0&1&0 \end{bmatrix}

    Finally, add (2nd row \times -\tfrac{1}{4}) to 1st row:
    \begin{bmatrix} 1&0&0&0&0\\0&1&-4&0&0\\0&0&0&1&0\\0&0&0&1&0\\0&0&0&1&0 \end{bmatrix}

    This tells you that:
    x_1 = 0
    x_2 - 4x_3 = 0
    x_4 = 1.

    Thus, x_1=0,x_4=1,x_2 = 4x_3 (notice that x_3 is "free").
    This means if t\in \mathbb{R} then \bold{x} = \begin{bmatrix}0\\4t\\t\\1 \end{bmatrix} solves M\bold{x} = \bold{0}, and conversely.

    Thus, the nullspace is \left\{ \begin{bmatrix}0\\4t\\t\\1 \end{bmatrix} : t\in \mathbb{R} \right\} = \left\{ t \begin{bmatrix}0\\4\\1\\1 \end{bmatrix} : t\in \mathbb{R}\right\}

    We see the dimension is one, and so the nullity is one.
    Ok I checked my working again, and it turns out i was wrong with reducing. The second line is not [0 1 -4 -2] but is [0 1 -4 6]. What does that change?
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  6. #6
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    rref(A) =
    1 0 1 -2
    0 1 -4 6
    0 0 0 0
    0 0 0 0
    0 0 0 0

    rank(A)=2, nullity(A)=2

    ker(A) = span(
    -1 2
    4 -6
    1 0
    0 1
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