
#1
Apr3010, 10:40 AM

P: 37

L: R^4 => R^3 is defined by L(x,y,z,w) = (x+y, z+w, x+z)
A) Find a basis for ker L We can re write L(x,y,z,w) as x* (1,01) + y *(1,0,0) + z*(0,1,1) + w*(0,1,0). I then reduced it to row echelon form We now have the equations XW=0 , Y+W=0, Z+W=0. There are infinitely many solutions as X=W, Y= W and Z=W. So if we set W=1 we have the basis for the kernel=Vector(1,1, 1,1) B) find a basis for range L Given L(x,y,z,w) = (x+y, z+w, x+z) We can re write L(x,y,z,w) as x* (1,01) + y *(1,0,0) + z*(0,1,1) + w*(0,1,0). S= {(1,01) ,(1,0,0) ,(0,1,1),(0,1,0)} It spans L. To find the basis for L we set {x* (1,01) + y *(1,0,0) + z*(0,1,1) + w*(0,1,0)} = 0,0,0 I reduced it and the leading one's appear in the first 3 columns of the reduced form, the first 3 vectors in the original matrix became a basis for the range of L They are: Vector( {(1, (0, (1}) , Vector( 1, 0, 0}) , Vector( 0, 0, 1}) C) verify theorem 10.7 (dim(KerL) + dim(rangeL) = dim V The dim can be viewed as the # of vectors in of the Ker/range. Given (dim(KerL) + dim(rangeL) = dim V we have 1+3=4, which is the number of dimensions in the original space (L(x,y,z,w)). 



#2
Apr3010, 01:39 PM

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P: 21,069





#3
Apr3010, 02:55 PM

P: 37

I forgot to type it. I believe A and B to be correct, but is my explanantion in C suffice?




#4
Apr3010, 03:23 PM

Mentor
P: 21,069

Kernel/Range basis
Sure, it's fine, and the other parts are fine also.



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