kernel and range

Upload: chatriaayon

Post on 13-Jan-2016

216 views

Category:

Documents


0 download

DESCRIPTION

teorema de la dimension

TRANSCRIPT

  • Kernel and Range

    When we talk about the kernel, we are asking for all the vectors in Rn that map to the zero vector in Rm. The range is all the vectors get in Rm. Examples: T: R2 R2, T(x,y) = (x 2y, 3x 6y) Kernel:

    To find the kernel, first get the standard matrix associated with the linear transformation. To get the standard matrix of this example, compute T(e1) = T(1,0) and T(e2) = T(0,1).

    T(e1) = T(1,0) = (1,3) T(e2) = T(0,1) = (-2,-6)

    Now the standard matrix [T] is given by

    63

    21.

    Next, use Gauss-Jordan to find the solution.

    =

    =

    =

    =

    ty

    tx

    ty

    yxRR

    202

    00

    21

    63

    21132

    So the kernel of this transformation is given by )2,1()2,1()ker( == tT , where t = 1.

    Range: To find the range, just take the columns of [T]. Thus the range is given by

    =

    6

    2,

    3

    1)(Trange

    T: R3 R2, T(x,y,z) = (x y + z, 2x + y z) Kernel: Again, to find the kernel we need to find the standard matrix and then put it reduced row-echelon form. T(e1) = T(1,0,0) = (1,2)

    T(e2) = T(0,1,0) = (-1,1) T(e3) = T(0,0,1) = (1, -1)

    Standard matrix [T]:

    112

    111

  • Use Gauss-Jordan gives us the following solution:

    =

    =

    =

    =

    =

    =

    +

    tz

    ty

    x

    tz

    zy

    x

    RRR

    RR

    0

    0

    0

    110

    001

    110

    111

    330

    111

    112

    11121

    23

    1

    122

    So the kernel of this transformation is given by: )1,1,0()1,1,0()ker( == tT , where t = 1.

    Range: Again to find the range, just take the columns of [T].

    =

    1

    1,

    1

    1,

    2

    1)(Trange

    Now, we can use the kernel and range to determine their dimension. dim(ker(T)) = nullity(T) = number of free variables

    dim(range(T)) = rank(T) = number of leading ones in reduced row-echelon form Example:

    512

    513

    311

    To find the dimension of the kernel associated with this matrix, we use Gauss-Jordan.

    =

    =

    =

    =

    =+

    =+

    +

    tz

    ty

    tx

    tz

    zy

    zxRRRR

    R

    RRRRRR

    2

    0

    02

    000

    110

    201

    440

    110

    311

    440

    110

    311

    110

    440

    311

    512

    513

    311

    42321

    2

    32123132

    So the dimension of the dim(ker(A)) = 1 and dim(range(T)) = 2. In general there is a formula relating the number of columns of [T] to the dimension of the kernel and the dimension of the range.

    dim(ker(T)) + dim(range(T)) = number of columns in [T]