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Math Help - Basis of kernel(T) where T is a linear transformation

  1. #1
    Senior Member x3bnm's Avatar
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    Basis of kernel(T) where T is a linear transformation

    Problem:
    I can't find answer to a problem. The problem is like this:

    Find the basis for \text{kernel of}(T) where T is a linear transformation.

    Suppose T:R^3 \to R^3,\quad T(x,y,z) = (x + 2y, y + 2z, z + 2x)

    Part of Solution:
    The problem is solved like this:

     A = \left[{\begin{array}{ccc} 1  & 2 & 0\\ 0 & 1 & 2\\ 2 & 0 & 1 \end{array}}\right]

    The reduced row echelon form of augmented matrix [A\vdots \mathbf{0}] is:

      \left[{\begin{array}{cccc} 1  & 0 & 0 & 0\\ 0 & 1 & 0 & 0\\ 0 & 0 & 1 & 0 \end{array}}\right]

    So the solutions are:

    x_1 = 0, x_2 = 0, x_3 = 0


    My question:

    How do you find the basis for ker(T) now because x_1 = x_2 = x_3 = 0?
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  2. #2
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    Re: Basis of kernel(T) where T is a linear transformation

    Quote Originally Posted by x3bnm View Post
    Problem:
    I can't find answer to a problem. The problem is like this:

    Find the basis for \text{kernel of}(T) where T is a linear transformation.

    Suppose T:R^3 \to R^3,\quad T(x,y,z) = (x + 2y, y + 2z, z + 2x)

    Part of Solution:
    The problem is solved like this:



     A = \left[{\begin{array}{ccc} 1  & 2 & 0\\ 0 & 1 & 2\\ 2 & 0 & 1 \end{array}}\right]

    The reduced row echelon form of augmented matrix [A\vdots \mathbf{0}] is:

      \left[{\begin{array}{cccc} 1  & 0 & 0 & 0\\ 0 & 1 & 0 & 0\\ 0 & 0 & 1 & 0 \end{array}}\right]

    So the solutions are:

    x_1 = 0, x_2 = 0, x_3 = 0


    My question:

    How do you find the basis for ker(T) now?
    You need to make your vectors the columns and not the rows!!

    Your proceedure is correct but you have used the wrong matrix.
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  3. #3
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    Re: Basis of kernel(T) where T is a linear transformation

    But you are still going to get the same thing: T(x, y, z)= (x+ 2y, y+ 2z, z+ 2x)= (0, 0, 0). That is, we must have x+ 2y= 0, y+ 2z= 0, and z+ 2x= 0. From the first equation, x= -2y. From the second, y= -2z so x= -2(-2z)= 4z. But from the third equation z= -2x so that x= 4(-2x)= -8x which is only satified for x= 0. Then z= -2x= 0 and y= -2z= 0. The entire kernel is the single 0 vector. That does not have a basis.
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  4. #4
    Senior Member x3bnm's Avatar
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    Re: Basis of kernel(T) where T is a linear transformation

    Quote Originally Posted by TheEmptySet View Post
    You need to make your vectors the columns and not the rows!!

    Your proceedure is correct but you have used the wrong matrix.
    Sorry I didn't understand what you said. Can you kindly explain a little? Sorry about that.
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  5. #5
    Senior Member x3bnm's Avatar
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    Re: Basis of kernel(T) where T is a linear transformation

    Quote Originally Posted by HallsofIvy View Post
    But you are still going to get the same thing: T(x, y, z)= (x+ 2y, y+ 2z, z+ 2x)= (0, 0, 0). That is, we must have x+ 2y= 0, y+ 2z= 0, and z+ 2x= 0. From the first equation, x= -2y. From the second, y= -2z so x= -2(-2z)= 4z. But from the third equation z= -2x so that x= 4(-2x)= -8x which is only satified for x= 0. Then z= -2x= 0 and y= -2z= 0. The entire kernel is the single 0 vector. That does not have a basis.
    Thanks HallsofIvy. That makes sense. One quick question.

    Can \left( \left[{\begin{array}{ccc} 0\\ 0\\ 0 \end{array}}\right]\right) ever be a basis? Is it possible? Is it valid? Again thanks for answering my question.
    Last edited by x3bnm; November 16th 2011 at 01:45 PM.
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  6. #6
    Senior Member x3bnm's Avatar
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    Re: Basis of kernel(T) where T is a linear transformation

    Quote Originally Posted by x3bnm View Post
    Can \left( \left[{\begin{array}{ccc} 0\\ 0\\ 0 \end{array}}\right]\right) ever be a basis? Is it possible? Is it valid? Again thanks for answering my question.
    I found the answer.

    Physics Forums - View Single Post - a question about (0,0,0) vector

    HallsofIvy Thanks again.
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