Working with linear operators

The linear operator T({a,b},{c,d}) = a+d.
a) Show that T is linear
b) Determine the dimension of kerT and give a basis of kerT
c) Find the basis for a complement to kerT
d) Putting together the basis for kerT and the complement found in the previous parts we get a basis of Mat(2x2,C). Find the matrix representation of T with respect to that basis.

Help would be much appreciated, I never have any idea what's going on in these lectures, I find the lecturer really hard to understand.

I hope I understand your notation. The argument inside of T is the 2x2 matrix

[a b]
[c d].

I'll denote this as {{a, b}, {c, d}}.

If this is correct, then T is a linear transformation from the space of 2x2 matrices with complex entries to the complex numbers. (This is not actually an operator since the domain and co-domain are different spaces.)

a) T is linear. Let {{a, b}, {c,d}} and {{x, y}, {z, w}} be any 2 2x2 complex matrices and let k be any scalar.

T({{a, b}, {c,d}} + {{x, y}, {z, w}}) = T({{a + x, b + y}, {c + z, d + w}}) = a + x + d + w

= (a + d) + (x + w) = T({{a, b}, {c,d}}) + T({{x, y}, {z, w}}).

Similarly

T(k{{a, b}, {c,d}}) = T({{ka, kb}, {kc,kd}}) = ka + kd = k(a + d) = k T({{a, b}, {c,d}}).

These are the two required properties for T to be linear.

b) Suppose that T(A) = 0, then for A = {{a, b}, {c,d}}, we clearly require that a + d = 0 ==> a = -d. There are no conditions on b and c. This means there are three "degrees of freedom"---we can have any a, any b, and any c, but d depends on a.

dim(ker(T)) = 3.

A basis can be constructed by setting one of a, b, and c to 1 and the others to zero (each in turn.)

B1 = {{1, 0}, {0, -1}}, B2 = {{0, 1}, {0, 0}}, and B3 = {{0, 0}, {1, 0}}.

A basis is B = {B1, B2, B3}. This is not the only possible basis, but it is any easy one to construct.