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Math Help - vectors that span the kernel of a matrix

  1. #1
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    vectors that span the kernel of a matrix

    Find vectors that span the kernel of matrix A.

    A = [1 2 3 4 ; 0 1 2 3 ; 0 0 0 1] where ; is a new row.

    I found that rref(A) = [1 0 -1 0 ; 0 1 2 0 ; 0 0 0 1] where ; is a new row.

    Can anyone tell me where I go from here? Thanks
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  2. #2
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    Solution space of AX = 0 will be equal to the the kernel of the mapping represented by A

    If you solve the equation we get the solution space is of dimension 1 with basis = [1 -2 1 0]
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  3. #3
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    Quote Originally Posted by noles2188 View Post
    Find vectors that span the kernel of matrix A.

    A = [1 2 3 4 ; 0 1 2 3 ; 0 0 0 1] where ; is a new row.

    I found that rref(A) = [1 0 -1 0 ; 0 1 2 0 ; 0 0 0 1] where ; is a new row.

    Can anyone tell me where I go from here? Thanks
    A= \begin{bmatrix}1 & 2 & 3 & 4 \\ 0 & 1 & 2 & 3\\ 0 & 0 & 0 & 1\end{bmatrix}.
    and
    rref(A)= \begin{bmatrix}1 & 0 & -1 & 0 \\ 0 & 1 & 2 & 0\\ 0 & 0 & 0 & 1\end{bmatrix}.
    Actually you can work from either A itself of rref(A). The "kernel" of a matrix is the subspace of all vectors that it maps to 0. So if v= \begin{bmatrix}a \\ b\\ c\\ d\end{bmatrix}is a vector in the kernel, we must have A(v)= \begin{bmatrix}1 & 2 & 3 & 4 \\ 0 & 1 & 2 & 3\\ 0 & 0 & 0 & 1\end{bmatrix}\begin{bmatrix}a \\ b\\ c\\ d\end{bmatrix}= \begin{bmatrix}0 \\ 0 \\ 0 \\ 0 \end{bmatrix}.

    That is the same as the system of equation a+ 2b+ 3c+ 4d= 0, b+ 2c+ 3d= 0, d= 0. You could, of course, reduce those to a- c= 0, b+ 2c= 0, d= 0, the row reduced form. From the first equation, a= c. From the second b= -2c. c can by any number, d must be 0. So any vector satisfying those equations, any vector in the kernel of A, must be of the form \begin{bmatrix}c \\ -2c\\ c\\ 0\end{bmatrix}= c\begin{bmatrix}1 \\ -2 \\ 1 \\ 0\end{bmatrix}.

    A basis for the kernel consists of that single vector \begin{bmatrix}1 \\ -2 \\ 1 \\ 0\end{bmatrix}, as aman_cc says.
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