Results 1 to 4 of 4

Thread: Kernel and range of a linear Transformation

  1. #1
    Newbie
    Joined
    Jul 2010
    Posts
    5

    Kernel and range of a linear Transformation

    $\displaystyle \ L:M_{22} \to M_{22}\ $ be defined by $\displaystyle
    \ L(A)=\begin{bmatrix}
    1 &2 \\
    1 & 1
    \end{bmatrix}A-A\begin{bmatrix}
    1 &2 \\
    1 & 1
    \end{bmatrix}\
    $

    Find the basis for ker L and basis for range L

    My work:
    --->
    $\displaystyle
    \ \ker L = \left \{ {L(A)| \begin{bmatrix}
    1 &2 \\
    1 & 1
    \end{bmatrix}}A-A\begin{bmatrix}
    1 &2 \\
    1 & 1
    \end{bmatrix}=\vec{0} \right \}\

    $

    $\displaystyle
    \[\ker L = \begin{bmatrix}
    1 &2 \\
    1 & 1
    \end{bmatrix}\begin{bmatrix}
    a &b \\
    c&d
    \end{bmatrix}-\begin{bmatrix}
    a &b\\
    c& d
    \end{bmatrix}\begin{bmatrix}
    1 &2 \\
    1 & 1
    \end{bmatrix}=\vec{0}\\
    =\begin{bmatrix}
    a+c \\
    b+d
    \end{bmatrix} - \begin{bmatrix}
    a+c \\
    b+d
    \end{bmatrix}=\begin{bmatrix}
    0&0 \\
    0&0
    \end{bmatrix}\\
    =...?\]
    \ $

    Sorry, I don't know how to find the kernel for an Mnn --> Mnn with an expression like this Can someone please help me out with this. Thanks in advance
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Senior Member roninpro's Avatar
    Joined
    Nov 2009
    Posts
    485
    You are fine all the way up to here:

    $\displaystyle \[\ker L = \begin{bmatrix}
    1 &2 \\
    1 & 1
    \end{bmatrix}\begin{bmatrix}
    a &b \\
    c&d
    \end{bmatrix}-\begin{bmatrix}
    a &b\\
    c& d
    \end{bmatrix}\begin{bmatrix}
    1 &2 \\
    1 & 1
    \end{bmatrix}=\vec{0}$

    The step after this does not make sense. Check your matrix multiplication!
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Jul 2010
    Posts
    5
    I apologize. I got pretty sloppy there. I was doing that late at night and so badly wanted to crash. Ok Here is what I got so far. Hopefully there wont typos...

    $\displaystyle
    \ L(A)=\begin{bmatrix}
    1 &2 \\
    1&1
    \end{bmatrix}\begin{bmatrix}
    a &b \\
    c & d
    \end{bmatrix}-\begin{bmatrix}
    a & b\\
    c& d
    \end{bmatrix}\begin{bmatrix}
    1& 2\\
    1 & 1
    \end{bmatrix}=
    \begin{bmatrix}
    a+2c & b+2b\\
    a+c & b+d
    \end{bmatrix}-\begin{bmatrix}
    a+b &2a+b \\
    c+d&2c+d
    \end{bmatrix}=\begin{bmatrix}
    0& 0\\
    0&0
    \end{bmatrix}\
    $

    $\displaystyle \ L(A)=\begin{bmatrix}
    2c-b & 2d-2a\\
    a+d & b-2c
    \end{bmatrix}=\begin{bmatrix}
    0& 0\\
    0&0
    \end{bmatrix}\ $

    I turned this into a matrix and performed rref

    $\displaystyle \ \begin{bmatrix}
    0 &-1 &2 & 0 &0 \\
    -2&0 &0 &2 &0 \\
    1& 0& 0 &1 &0 \\
    0&1 & -2 & 0 &0
    \end{bmatrix}\sim \begin{bmatrix}
    1 &0 &0 & 0 &0 \\
    0&1 &-2&0 &0 \\
    0& 0& 0 &1 &0 \\
    0&0 & 0 & 0 &0
    \end{bmatrix}\ $

    a= 0
    b = 2
    c = arbitrary
    d - 0

    However the correct answer for the basis for ker L is:
    $\displaystyle \ \left \{ \begin{bmatrix}
    1&0 \\
    0&1
    \end{bmatrix} ,\begin{bmatrix}
    0 &1 \\
    1/2& 0
    \end{bmatrix}\right \}\ $

    I don't know how to answer this.
    I also have to find the basis for range L.
    The correct answer for the basis for range L is:
    $\displaystyle \ \left \{ \begin{bmatrix}
    0&-2 \\
    1&0
    \end{bmatrix} ,\begin{bmatrix}
    -1 &0\\
    0& 1
    \end{bmatrix}\right \}\ $
    But I don't know how to get that since I don't know how to get the ker L for this problem. I know that range L = span S. Not sure how to set this up....
    Can someone please help me figure out this problem. Thanks in advance.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor

    Joined
    Apr 2005
    Posts
    19,775
    Thanks
    3028
    Quote Originally Posted by wilday86 View Post
    I apologize. I got pretty sloppy there. I was doing that late at night and so badly wanted to crash. Ok Here is what I got so far. Hopefully there wont typos...

    $\displaystyle
    \ L(A)=\begin{bmatrix}
    1 &2 \\
    1&1
    \end{bmatrix}\begin{bmatrix}
    a &b \\
    c & d
    \end{bmatrix}-\begin{bmatrix}
    a & b\\
    c& d
    \end{bmatrix}\begin{bmatrix}
    1& 2\\
    1 & 1
    \end{bmatrix}=
    \begin{bmatrix}
    a+2c & b+2b\\
    a+c & b+d
    \end{bmatrix}-\begin{bmatrix}
    a+b &2a+b \\
    c+d&2c+d
    \end{bmatrix}=\begin{bmatrix}
    0& 0\\
    0&0
    \end{bmatrix}\
    $

    $\displaystyle \ L(A)=\begin{bmatrix}
    2c-b & 2d-2a\\
    a+d & b-2c
    \end{bmatrix}=\begin{bmatrix}
    0& 0\\
    0&0
    \end{bmatrix}\ $
    The lower left entry should be "a- d", not "a+ d".
    Rather than go to 5 by 4 matrices, note that this says the 2c- b= 0, 2d- 2a= 0, a- d= 0, and b- 2c= 0. The second and third equations both give a= d. From the first equation b= 2c so the fourth equation becomes 2c- 2c= 0 which is true for all c. These will be true for all d= a and b=2c: $\displaystyle \begin{bmatrix}a & 2c \\ c & a\end{bmatrix}= a\begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix}+ c\begin{bmatrix}0 & 2 \\ 1 & 0\end{bmatrix}$.

    I turned this into a matrix and performed rref

    $\displaystyle \ \begin{bmatrix}
    0 &-1 &2 & 0 &0 \\
    -2&0 &0 &2 &0 \\
    1& 0& 0 &1 &0 \\
    0&1 & -2 & 0 &0
    \end{bmatrix}\sim \begin{bmatrix}
    1 &0 &0 & 0 &0 \\
    0&1 &-2&0 &0 \\
    0& 0& 0 &1 &0 \\
    0&0 & 0 & 0 &0
    \end{bmatrix}\ $

    a= 0
    b = 2
    c = arbitrary
    d - 0

    However the correct answer for the basis for ker L is:
    $\displaystyle \ \left \{ \begin{bmatrix}
    1&0 \\
    0&1
    \end{bmatrix} ,\begin{bmatrix}
    0 &1 \\
    1/2& 0
    \end{bmatrix}\right \}\ $
    Of course $\displaystyle \begin{bmatrix}0 & 2 \\ 1 & 0\end{bmatrix}= 2\begin{bmatrix}0 & 1 \\\frac{1}{2} & 0\end{bmatrix}$ so this is just a variation on my answer.

    I don't know how to answer this.
    I also have to find the basis for range L.
    The correct answer for the basis for range L is:
    $\displaystyle \ \left \{ \begin{bmatrix}
    0&-2 \\
    1&0
    \end{bmatrix} ,\begin{bmatrix}
    -1 &0\\
    0& 1
    \end{bmatrix}\right \}\ $
    But I don't know how to get that since I don't know how to get the ker L for this problem. I know that range L = span S. Not sure how to set this up....
    Can someone please help me figure out this problem. Thanks in advance.
    range L= span S? You haven't said what S is.

    The range of L is all matrics [tex]\begin{bmatrix}u & v \\ w & x\end{bmatrix} such that $\displaystyle \begin{bmatrix}u & v \\ w & x\end{bmatrix}= L(A)= \begin{bmatrix}2c- b & 2d- 2a \\ a- d & b- 2c\end{bmatrix}$ for some a, b, c, d. We need to reduce u= 2c- b, v= 2d- 2a, w= a- d, and x= b- 2c to equations in u, v, w, and x only. Obviously the second and third equation say v= 2w. From the first equation b= 2c- u and then the fourth equation becomes x= (2c-u)- 2c= -u. We can write any matrix in the range of L as $\displaystyle \begin{bmatrix}u & 2w \\ w & -u\end{bmatrix}= u\begin{bmatrix}1 & 0 \\ 0 & -1\end{bmatrix}+ w\begin{bmatrix}0 & 2 \\ 1 & 0\end{bmatrix}$ which is equivalent to the basis you give.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. [SOLVED] Basis of kernel(T) where T is a linear transformation
    Posted in the Advanced Algebra Forum
    Replies: 5
    Last Post: Nov 16th 2011, 12:42 PM
  2. Finding a Linear Transformation given its Kernel
    Posted in the Advanced Algebra Forum
    Replies: 7
    Last Post: Feb 2nd 2011, 10:06 PM
  3. Kernel and range of linear transformation
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: Apr 13th 2009, 06:26 PM
  4. Kernel and Range of Linear Transformation
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: Mar 29th 2009, 08:02 AM
  5. Kernel and Range of a linear operator
    Posted in the Advanced Algebra Forum
    Replies: 0
    Last Post: Mar 7th 2009, 11:23 AM

Search Tags


/mathhelpforum @mathhelpforum