pure maths -matrix

i cannot post a picture as solution.

Ai) obvious, just put everything at L.H.S, than try to re-express it in the form of R.H.S.

Aii) Expend LHS, and RHS, and you will found they equal to same thing.

B) the equation A*B=[0 0] (* is the triangle)

and think about the B is the solution but not zero.

therefore, the determine of [a1 -a2 ; a2 a1] = 0
i.e.
a1^2+a2^2=0 => a1 and a2=0

thus, A=[0 0]

Ci) since any matrix X = [x y] = x E1+ y E2, where x,y belong real
by property 1, T(X)=T (x E1+ y E2)= x T(E1)+y T(E2)=x U*E1+y U*E2
by Ai) x U*E1+y U*E2 = U*(x E1+y E2)=U*X

Cii) record the definition of surjective: every Y in M (from the codomain), there exist some X in M (from the domain), such that T(X)=Y

the transform T(x)=Y is look like U*X=Y (by Ci)
the equation U*X=Y, for any Y, because U is not zero, i.e, determent of U*X is not zero, thus, X has unique solution. so T is surjective.

i am sorry that i cannot give a good looking proof. but i think you better work it out by yourself :D i just give a guide line.