You are fine all the way up to here:
The step after this does not make sense. Check your matrix multiplication!
be defined by
Find the basis for ker L and basis for range L
My work:
--->
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
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...
I turned this into a matrix and performed rref
a= 0
b = 2
c = arbitrary
d - 0
However the correct answer for the basis for ker L is:
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:
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.
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: .
Of course so this is just a variation on my answer.I turned this into a matrix and performed rref
a= 0
b = 2
c = arbitrary
d - 0
However the correct answer for the basis for ker L is:
range L= span S? You haven't said what S is.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:
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.
The range of L is all matrics [tex]\begin{bmatrix}u & v \\ w & x\end{bmatrix} such that 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 which is equivalent to the basis you give.