# Kernel and range of a linear Transformation

• July 25th 2010, 08:05 PM
wilday86
Kernel and range of a linear Transformation
$\ L:M_{22} \to M_{22}$/extract_itex] be defined by $ \ 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: ---> $ \ \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 \}\ $ $ \[\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(Doh) Can someone please help me out with this. Thanks in advance
• July 25th 2010, 10:57 PM
roninpro
You are fine all the way up to here:

$\[\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!
• July 26th 2010, 06:38 PM
wilday86
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...

$
\ 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}\
$

$\ 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

$\ \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:
$\ \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. (Worried)
I also have to find the basis for range L.
The correct answer for the basis for range L is:
$\ \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....
• July 27th 2010, 04:35 AM
HallsofIvy
Quote:

Originally Posted by wilday86
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...

$
\ 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}\
$

$\ 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: $\begin{bmatrix}a & 2c \\ c & a\end{bmatrix}= a\begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix}+ c\begin{bmatrix}0 & 2 \\ 1 & 0\end{bmatrix}$.

Quote:

I turned this into a matrix and performed rref

$\ \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:
$\ \left \{ \begin{bmatrix}
1&0 \\
0&1
\end{bmatrix} ,\begin{bmatrix}
0 &1 \\
1/2& 0
\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.

Quote:

I don't know how to answer this. (Worried)
I also have to find the basis for range L.
The correct answer for the basis for range L is:
$\ \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....
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.