M2(R) is the vector space of 2x2 matrices with real entries where addition and scaling are done componentwise. There are several similar notations for similar notions. As I mentioned ker T (read "the kernel of T") is the set of elements mapped to 0 by T. As your previous question showed, ker T is a vector space itself (being a subspace of T's domain), so "taking the kernel" of a linear transformation is one nice way to generate new vector (sub)spaces. I showed that ker T always contains 0, but perhaps ker T is trivial and contains nothing more. A key question along these lines, then, is whether or not the kernel of a given linear transformation T contains more than just the zero vector. (The same type of reasoning pervades abstract algebra in more general contexts.)

(a) T(1,2,3,4) is the matrix {{1+2*2-3, 2}, {-2, 4-1}} = {{2, 2}, {-2, 3}}. The zero element of M2(R) is of course just the 0 matrix, {{0, 0}, {0, 0}}. Note that T is indeed a linear transformation (I will not verify it). To get T(a, b, c, d) to be {{0, 0}, {0, 0}}, we need...

a + 2b - c = 0
b = 0
-b = 0
d-a = 0

while not all of a, b, c, and d are 0. Solving the system above gives solutions if and only if
b=0, a=c=d
so (1,0,1,1) qualifies. Indeed (r, 0, r, r) is the form of all solutions--and the kernel is the one dimensional vector space {(r, 0, r, r) where r is real}.

(b) im T is the image of T, i.e. the set of all elements in T's codomain that are actually hit by applying T to some element of its domain. It's very similar to ker T in that this operation generates another subspace (proof omitted) and has general uses in algebra. Determining a basis for the image implicitly is extremely simple, since...
T(w, x, y, z) = w T(1, 0, 0, 0) + x T(0, 1, 0, 0) + y T(0, 0, 1, 0) + z T(0, 0, 0, 1)

That is, every element T(w, x, y, z) in im T is a linear combination of {T(e1), T(e2), T(e3), T(e4)} where the ei are V4's standard basis vectors. However, if for instance, T(e1) = T(e2) = 0, we could remove T(e1) and T(e2) from this list, so in general we must remove such redundancy--we must remove enough elements so that the resulting set is linearly independent (in general removing 0 or more vectors; which ones we remove turns out to be relevant). Here, the above (multi)set is

{(1, 1, 1), (-1, 0, 1), (1, 2, 3), (1, -1, -3)}

Removing linear dependence is in general somewhat difficult though in this case is easy--after our kernel computation, we will find that the basis has length 2, so we must remove 2 of the above elements, and which two is irrelevant in this particular case since any two of them are together obviously linearly independent. Answers are {T(e1), T(e2)}, {T(e1), T(e3)}, etc.


Computing a basis for ker T:
T(w,x,y,z) = (0,0,0)

iff

w-x+y+z = 0
w+2y-z = 0
w+x+3y-3z = 0

iff

z = w+2y
x = w+y+(w+2y) = 2w + 3y
w + (2w+3y) + 3y - 3(w+2y) = 0, i.e. 0=0

iff

z = w+2y
x = 2w+3y

Vectors in the kernel are then of the form (w, 2w+3y, y, w+2y) = w(1, 2, 0, 1) + y(0, 3, 1, 2) for w and y arbitrary, so the basis is just {(1, 2, 0, 1), (0, 3, 1, 2)} (note these are linearly independent). The rank-nullity theorem can be phrased as: for a linear transformation T, the dimension of the kernel and the dimension of the image sum to the dimension of the domain, i.e. dim(ker T) + dim(im T) = dim(dom T). (dim(ker T) is termed the "nullity", perhaps since ker T is termed "the null space"; dim(im T) is termed the "rank" of T, hence the name of the theorem.) Since dim(ker T) = 2 and dim(V4) = 4, we must have dim(im T) = 4 - 2 = 2. This justifies my comments at the end of the previous half of this problem.

(c) Given a matrix A, its characteristic polynomial is det(A - tI) where t is an indeterminate and I is the appropriately sized identity matrix. A has eigenvalues precisely at the roots of its characteristic polynomial. Usually characteristic polynomials are considered to be polynomials in a real variable (as in the previous sentence), but it is also possible to substitute in matrices in place of "t" (after computing the determinant in the usual way); the constant term is interpreted as multiplied by the appropriate identity matrix. Here...

det(A - tI) = det({{1-t, 3}, {4, 5-t}})
= (1-t)(5-t) - 3*4
= t^2 - 6t - 7

Plugging in t=A gives...
A^2 - 6A - 7(I)
= {{1, 3}, {4, 5}}{{1, 3}, {4, 5}} - 6{{1, 3}, {4, 5}} - {{7, 0}, {0, 7}}
= {{1+12, 3+15}, {4+20}, {12+25}} - {{6+7, 18}, {24, 30+7}}
= {{13-13, 18-18}, {24-24, 37-37}}
= {{0, 0}, {0, 0}}

so A is indeed a "root" in this sense. This is a special case of the Cayley-Hamilton theorem.