# Kernal and Range

• Dec 20th 2012, 02:02 AM
ohYeah
Kernal and Range
HI I am studying for the final which is in less then 7 hours, There is this one question on the review sheet which I really can't do at all, for some reason I just think I can't do Kernal and Range.
I looked at the examples in the text book all 23 of them none are similar to these that way i can just repeat the sets and follow.

so please if anyone can show me how to do these with clear steps, you will have saved me in the final. This website has already taught me more then my class in Linear. But this is the final so please your help will be more then appreciated. There is no one left to go and as

here is the question
Attachment 26309
• Dec 20th 2012, 03:11 AM
Deveno
Re: Kernal and Range
for a linear transformation T:

kernel = which vectors in the domain (the one of the left side of the arrow in T:U→V, that is: U) T takes to the 0-vector of the co-domain (that is: V).

range = which vectors in the co-domain that are images under T: that is those v in V for which there is a u in U with T(u) = v.

the kernel lives in U, the range lives in V.

problem (a):

if T is linear, we have to have T(mA) = mT(A) for EVERY nxn matrix A. well, the nxn identity matrix I is an nxn matrix, is it true that det(mI) = m(det(I)) (think carefully before you answer)?

problem (b): surely you know by now that multiplication by a matrix is a linear map. what else have you been doing in this course?

let's look at what TA does to our favorite basis vectors: (1,0,0), (0,1,0) and (0,0,1). oh look! these are just the columns of the matrix:

(3,-1), (-5,2) and (2,4).

well, R2 only has dimension 2, so we can't have a basis with 3 things in it.

so (2,4) must be realizable as a linear combination of (3,-1) and (-5,2):

a(3,-1) + b(-5,2) = (2,4) that is:

(3a-5b,2b-a) = (2,4), or in "table form":

3a-5b = 2
2b-a = 4 which is the same as:

3a-5b = 2
-3a+6b = 12

adding, we get: b = 14, so a = 24. checking: 24(3,-1) + 14(-5,2) = (72,-24) + (-70,28) = (2,4).

so span({(3,-1),(-5,2),(2,4)}) = span({(3,-1),(-5,2)}). thus dim(range(TA)) ≤ 2.

(well, of course it is, dim(R) = 2, and this is a subsapce!). the real question is: are (3,-1) and (-5,2) linearly independent?

ok, suppose c1(3,-1) + c2(-5,2) = (0,0).

this means (3c1-5c2,2c2-c1) = (0,0) (this should start to seem familiar, now....)

so:

3c1-5c2 = 0
2c2-c1 = 0

multiplying the second equation by 3, we get:

3c1-5c2 = 0
-3c1+6c2 = 0

adding, we get: c2 = 0. thus, c1 = 0. so {(3,-1),(-5,2)} is indeed a basis for range(TA). but since this has dimension 2 (2 elements in our basis), it must be ALL of R2 (which also has dimension 2).

now, by the rank-nullity theroem, we have:

3 = dim(ker(TA)) + dim(range(TA)) = dim(ker(TA) + 2.

hence dim(ker(TA)) = 1. and THIS means all we have to do is find ONE non-zero vector (x1,x2,x3) with T(x1,x2,x3) = (0,0).

well we saw before that 24T(1,0,0) + 14T(0,1,0) = T(0,0,1).

therefore, 24T(1,0,0) + 14T(0,1,0) - T(0,0,1) = (0,0), so:

T(24,0,0) + T(0,14,0) + T(0,0,-1) = (0,0) that is:

T(24,14,-1) = (0,0)<----(24,14,-1) is just such a non-zero vector. so {(24,14,-1)} is a basis for ker(TA).

for (c), we have to verify first that T is linear (it's not obvious).

this is for any two polynomials p(x) = ax2+bx+c and q(x) = dx2+ex+f, and any real number r, we need:

A) T(p(x) + q(x)) = T(p(x)) + T(q(x))
B) T(r(p(x)) = rT(p(x)).

note that p'(2) = 2a(2) + b = 4a + b, and q'(2) = 4d + e.

we'll verify (A) first:

T(p(x) + q(x)) = T(ax2+bx+c + dx2+ex+f) = T((a+d)x2+(b+e)x+(c+f)) = 4(a+d) + (b+e)

= 4a + b + 4d + e = p'(2) + q'(2) = T(p(x)) + T(q(x)).

now (B):

T(r(p(x)) = T(rax2+rbx+rc) = 4(ra) + rb = r(4a + b) = r(p'(2)) = r(T(p(x)), so T is indeed linear.

this time, let's focus on ker(T), first.

which polynomials p(x), do we have T(p(x)) = 0?

well T(p(x)) = p'(2) = 4a+b, so if this is 0, we need b = -4a. since the constant term goes away when we differentiate, c might be anything.

thus ker(T) is all polynomials of the form: p(x) = ax2 - 4ax + c = a(x2 - 4x) + c.

again, using the rank-nullity theorem, and the fact that dim(P2) = 3 (a basis is {x2,x,1}) we have:

3 = dim(ker(T)) + dim(range(T)). and dim(range(T)) ≤ 1, so dim(ker(T)) ≥ 2.

well, not EVERY element of P2 is in ker(T), for example T(x) = 1 ≠ 0 (the constant function 1(x) = 1, for all x is never 0).

so dim(ker(T)) is not 3, so it must be 2.

hmm.....do you suppose that {x2 - 4x,1} might be a basis?

are they linearly independent? suppose r(x2 - 4x) + s(1) = 0 (the 0-polynomial: 0x2+0x+0).

then rx2- 4rx + s = 0x2 + 0x + 0, so:

r = 0
-4r = 0
s = 0...so r and s must be 0, so our two polynomials are indeed linearly independent. thus we've found a basis for ker(T).

by rank-nullity, the range must have dimension 1. since R itself has dimension 1, the range must be all of R! this has basis {1}.

let's prove this by finding a p(x) with T(p(x)) = r, for any real-number r. since the constant term of our polynomial doesn't affect the derivative, let's suppose p(x) = ax2 + bx (0 constant term).

since T(p(x)) = p'(2) = 4a + b, let's choose a = 0, and see if we can find a b that works. hmm...it looks like b = r works.

so let p(x) = rx = 0x2 + rx + 0.

p'(x) = r, which is a constant function: r(x) = r, for all x.

so T(p(x)) = p'(2) = r(2) = r, showing that the range of T is indeed all of R.
• Dec 20th 2012, 01:33 PM
ohYeah
Re: Kernal and Range
Thank You for the thorough explanation, It took a while but I actually understand it now, and I know it must have taken a while to type all this up. I am really in dept to you for all you have done

and for a) its not a linear transformation because it did Fail this property "T(mA) = mT(A)" since the det(mA) ≠ m det (a) and i used a basic 2 x 2 matrix to prove that (1,0), (0,1) * 2 will equal (2,0),(0,2) the det of which is 4
but det(1,0), (0,1) = 1 * 2 = 2 ≠ 4

again thank you for eveyrthing