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Math Help - Finding the Kernel for a Transformation matrix

  1. #1
    lo2
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    Finding the Kernel for a Transformation matrix

    I have the following Transformation matrix F=\begin{bmatrix}1&1&1&0 \end{bmatrix} then I have to find the Kernel for it.

    I do that by solving the equation:

    \begin{bmatrix}1&1&1&0 \end{bmatrix} \cdot \begin{bmatrix}x_1\\x_2\\x_3\\x_4 \end{bmatrix}=0.

    But when I multiplicy these two matrices I get a 1x1 matrix.

    And I know that there ought to be 3 Kernel matrices (from Maple )?

    SO what do I do?
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  2. #2
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    Quote Originally Posted by lo2 View Post
    I have the following Transformation matrix F=\begin{bmatrix}1&1&1&0 \end{bmatrix} then I have to find the Kernel for it.

    I do that by solving the equation:

    \begin{bmatrix}1&1&1&0 \end{bmatrix} \cdot \begin{bmatrix}x_1\\x_2\\x_3\\x_4 \end{bmatrix}=0.

    But when I multiplicy these two matrices I get a 1x1 matrix.

    And I know that there ought to be 3 Kernel matrices (from Maple )?

    SO what do I do?
    This gives you,
    x_1+x_2+x_3=0 \implies x_1 = - x_2 - x_3.
    Now if you let x_4 = s,x_3=t,x_2=r then x_1 =  - r - t.

    Then the kernel is \{ (-r - t, r,t,s)|r,t,s\in \mathbb{R}\}
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  3. #3
    lo2
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    Quote Originally Posted by ThePerfectHacker View Post
    This gives you,
    x_1+x_2+x_3=0 \implies x_1 = - x_2 - x_3.
    Now if you let x_4 = s,x_3=t,x_2=r then x_1 =  - r - t.

    Then the kernel is \{ (-r - t, r,t,s)|r,t,s\in \mathbb{R}\}
    But when I use Maple to generate NullSpace I get 3 vectors:

    \begin{bmatrix}-1\\1\\0\\0 \end{bmatrix}, \begin{bmatrix}0\\0\\0\\1 \end{bmatrix}, \begin{bmatrix}-1\\0\\1\\0 \end{bmatrix}

    How come?

    And another question I have to find a Linear Transformation g: R^{2x2} \rightarrow R where U (the amount of 2x2 symmetric matrices where A=A^T). But How can I do that when kernel is a 1x1?
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    Quote Originally Posted by lo2 View Post
    But when I use Maple to generate NullSpace I get 3 vectors:

    \begin{bmatrix}-1\\1\\0\\0 \end{bmatrix}, \begin{bmatrix}0\\0\\0\\1 \end{bmatrix}, \begin{bmatrix}-1\\0\\1\\0 \end{bmatrix}

    How come?
    That is because Maple is giving you a basis for the nullspace. What we wrote above was correct. We shown that the nullspace is:
    \left\{ \begin{bmatrix} -r-t\\r\\t\\s \end{bmatrix}: r,s,t\in \mathbb{R} \right\}

    Now,
    \begin{bmatrix} -r-t\\r\\t\\s \end{bmatrix} = r\begin{bmatrix} -1\\1\\0\\0\end{bmatrix} + t \begin{bmatrix}-1\\0\\1\\0 \end{bmatrix} + s \begin{bmatrix} 0\\0\\0\\1 \end{bmatrix}

    Thus, we see \left\{ \begin{bmatrix} -1\\1\\0\\0\end{bmatrix}, \begin{bmatrix}-1\\0\\1\\0 \end{bmatrix}, \begin{bmatrix} 0\\0\\0\\1 \end{bmatrix} \right\} is a basis for the nullspace.
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  5. #5
    lo2
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    I would say that

    kerf = span{(-1, 1, 0, 0), (-1, 0, 1, 0), (0, 0, 0, 1)}

    is a satisfying answer. Agree?

    And another question I have to find a Linear Transformation g: R^{2x2} \rightarrow R where U (the amount of 2x2 symmetric matrices where A=A^T). But How can I do that when kernel is a 1x1??
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    Quote Originally Posted by lo2 View Post
    I would say that

    kerf = span{(-1, 1, 0, 0), (-1, 0, 1, 0), (0, 0, 0, 1)}

    is a satisfying answer. Agree?
    Yes

    And another question I have to find a Linear Transformation g: R^{2x2} \rightarrow R where U (the amount of 2x2 symmetric matrices where A=A^T). But How can I do that when kernel is a 1x1??
    I am confused here. By \mathbb{R}^{2\times 2} you mean the space of all 2\times 2 symmetric matrices? If so you can always let g: \mathbb{R}^{2\times 2}\to \mathbb{R} be the trivial mapping i.e. g(A) = 0. Then g is certainly a linear transformation.
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  7. #7
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    Quote Originally Posted by lo2 View Post
    I would say that

    kerf = span{(-1, 1, 0, 0), (-1, 0, 1, 0), (0, 0, 0, 1)}

    is a satisfying answer. Agree?

    And another question I have to find a Linear Transformation g: R^{2x2} \rightarrow R where U (the amount of 2x2 symmetric matrices where A=A^T). But How can I do that when kernel is a 1x1??
    Is there something missing after "where U"? Where U what? And I don't know what you mean by "kernel is a 1x1". The kernel of any Linear Transformation from vector space A to vector space B is a subspace of A.

    Are you required to find a transformation that maps 2 by two matrices to a single number and has "all symmetric matrices" as kernel? (You say "2x2 symmertric matrices where A= A^T but that is true of all symmetric matrices.)
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  8. #8
    lo2
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    Well I will try and explain it.

    I have this linear transformation g: R^{2x2} \rightarrow R. And I have this subspace U (a subspace of R^{2x2}) that is all 2x2 diagonal matrices where A=A^T.

    Then I have to find a linear transformation g: R^{2x2} \rightarrow R where the kernel is U.

    But U is a 2x2 matrix and a such transformation gives a 1x1 matrix...?

    What to do!?
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  9. #9
    lo2
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    Sorry but I really need help soon...!

    So I am kind of desperat and I would really like some help!
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  10. #10
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    Quote Originally Posted by lo2 View Post
    Well I will try and explain it.

    I have this linear transformation g: R^{2x2} \rightarrow R. And I have this subspace U (a subspace of R^{2x2}) that is all 2x2 diagonal matrices where A=A^T.

    Then I have to find a linear transformation g: R^{2x2} \rightarrow R where the kernel is U.

    But U is a 2x2 matrix and a such transformation gives a 1x1 matrix...?

    What to do!?
    So you want a linear transformation that maps precisely the symmetric matrices to 0. That is, you want a function that maps the matrix \left[\begin{array}{cc}a & b \\ c & d\end{array}\right] to 0 if and only if b= c. What about taking it to b- c? Is that a linear transformation?
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