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 T_{A}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, R^{2}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(T_{A})) ≤ 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 c_{1}(3,-1) + c_{2}(-5,2) = (0,0).

this means (3c_{1}-5c_{2},2c_{2}-c_{1}) = (0,0) (this should start to seem familiar, now....)

so:

3c_{1}-5c_{2}= 0

2c_{2}-c_{1}= 0

multiplying the second equation by 3, we get:

3c_{1}-5c_{2}= 0

-3c_{1}+6c_{2}= 0

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

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

3 = dim(ker(T_{A})) + dim(range(T_{A})) = dim(ker(T_{A}) + 2.

hence dim(ker(T_{A})) = 1. and THIS means all we have to do is find ONE non-zero vector (x_{1},x_{2},x_{3}) with T(x_{1},x_{2},x_{3}) = (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(T_{A}).

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

this is for any two polynomials p(x) = ax^{2}+bx+c and q(x) = dx^{2}+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(ax^{2}+bx+c + dx^{2}+ex+f) = T((a+d)x^{2}+(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(rax^{2}+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) = ax^{2}- 4ax + c = a(x^{2}- 4x) + c.

again, using the rank-nullity theorem, and the fact that dim(P_{2}) = 3 (a basis is {x^{2},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 P_{2}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 {x^{2}- 4x,1} might be a basis?

are they linearly independent? suppose r(x^{2}- 4x) + s(1) = 0 (the 0-polynomial: 0x^{2}+0x+0).

then rx^{2}- 4rx + s = 0x^{2}+ 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) = ax^{2}+ 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 = 0x^{2}+ 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.