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Math Help - Kernel and range of a linear Transformation

  1. #1
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    Kernel and range of a linear Transformation

     \ L:M_{22} \to M_{22}\ be defined by <br />
\ L(A)=\begin{bmatrix}<br />
1 &2 \\ <br />
1 & 1<br />
\end{bmatrix}A-A\begin{bmatrix}<br />
1 &2 \\ <br />
1 & 1<br />
\end{bmatrix}\ <br />

    Find the basis for ker L and basis for range L

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

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

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

    \[\ker L = \begin{bmatrix}<br />
1 &2 \\<br />
1 & 1<br />
\end{bmatrix}\begin{bmatrix}<br />
a &b \\<br />
c&d<br />
\end{bmatrix}-\begin{bmatrix}<br />
a &b\\<br />
c& d<br />
\end{bmatrix}\begin{bmatrix}<br />
1 &2 \\<br />
1 & 1<br />
\end{bmatrix}=\vec{0}

    The step after this does not make sense. Check your matrix multiplication!
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  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...

    <br />
\ L(A)=\begin{bmatrix}<br />
1 &2 \\ <br />
 1&1 <br />
\end{bmatrix}\begin{bmatrix}<br />
a &b \\ <br />
c & d<br />
\end{bmatrix}-\begin{bmatrix}<br />
a & b\\ <br />
 c& d<br />
\end{bmatrix}\begin{bmatrix}<br />
1& 2\\ <br />
1 & 1<br />
\end{bmatrix}=<br />
\begin{bmatrix}<br />
a+2c & b+2b\\ <br />
a+c & b+d<br />
\end{bmatrix}-\begin{bmatrix}<br />
a+b &2a+b \\ <br />
c+d&2c+d<br />
\end{bmatrix}=\begin{bmatrix}<br />
 0& 0\\ <br />
 0&0 <br />
\end{bmatrix}\<br />

    \ L(A)=\begin{bmatrix}<br />
2c-b & 2d-2a\\ <br />
a+d & b-2c<br />
\end{bmatrix}=\begin{bmatrix}<br />
 0& 0\\ <br />
 0&0 <br />
\end{bmatrix}\

    I turned this into a matrix and performed rref

    \ \begin{bmatrix}<br />
0 &-1  &2  & 0 &0 \\ <br />
 -2&0  &0  &2  &0 \\ <br />
 1&  0& 0 &1  &0 \\ <br />
 0&1  & -2 & 0 &0 <br />
\end{bmatrix}\sim \begin{bmatrix}<br />
1 &0 &0  & 0 &0 \\ <br />
 0&1 &-2&0  &0 \\ <br />
 0&  0& 0 &1  &0 \\ <br />
 0&0  & 0 & 0 &0 <br />
\end{bmatrix}\

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

    However the correct answer for the basis for ker L is:
    \ \left \{ \begin{bmatrix}<br />
 1&0 \\ <br />
 0&1 <br />
\end{bmatrix} ,\begin{bmatrix}<br />
0 &1 \\ <br />
 1/2& 0<br />
\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:
    \ \left \{ \begin{bmatrix}<br />
 0&-2 \\ <br />
 1&0 <br />
\end{bmatrix} ,\begin{bmatrix}<br />
-1 &0\\ <br />
 0& 1<br />
\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.
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  4. #4
    MHF Contributor

    Joined
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    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...

    <br />
\ L(A)=\begin{bmatrix}<br />
1 &2 \\ <br />
 1&1 <br />
\end{bmatrix}\begin{bmatrix}<br />
a &b \\ <br />
c & d<br />
\end{bmatrix}-\begin{bmatrix}<br />
a & b\\ <br />
 c& d<br />
\end{bmatrix}\begin{bmatrix}<br />
1& 2\\ <br />
1 & 1<br />
\end{bmatrix}=<br />
\begin{bmatrix}<br />
a+2c & b+2b\\ <br />
a+c & b+d<br />
\end{bmatrix}-\begin{bmatrix}<br />
a+b &2a+b \\ <br />
c+d&2c+d<br />
\end{bmatrix}=\begin{bmatrix}<br />
 0& 0\\ <br />
 0&0 <br />
\end{bmatrix}\<br />

    \ L(A)=\begin{bmatrix}<br />
2c-b & 2d-2a\\ <br />
a+d & b-2c<br />
\end{bmatrix}=\begin{bmatrix}<br />
 0& 0\\ <br />
 0&0 <br />
\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: \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

    \ \begin{bmatrix}<br />
0 &-1  &2  & 0 &0 \\ <br />
 -2&0  &0  &2  &0 \\ <br />
 1&  0& 0 &1  &0 \\ <br />
 0&1  & -2 & 0 &0 <br />
\end{bmatrix}\sim \begin{bmatrix}<br />
1 &0 &0  & 0 &0 \\ <br />
 0&1 &-2&0  &0 \\ <br />
 0&  0& 0 &1  &0 \\ <br />
 0&0  & 0 & 0 &0 <br />
\end{bmatrix}\

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

    However the correct answer for the basis for ker L is:
    \ \left \{ \begin{bmatrix}<br />
 1&0 \\ <br />
 0&1 <br />
\end{bmatrix} ,\begin{bmatrix}<br />
0 &1 \\ <br />
 1/2& 0<br />
\end{bmatrix}\right \}\
    Of course \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:
    \ \left \{ \begin{bmatrix}<br />
 0&-2 \\ <br />
 1&0 <br />
\end{bmatrix} ,\begin{bmatrix}<br />
-1 &0\\ <br />
 0& 1<br />
\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 \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 \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.
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