Well i will try to help... ( my english is bad so if you dont understand some words just tell and i will try to explain better )
In a) part you need to give a basis for V.
So the "best" basis for any space is canon basis
( example.
for space R^2 the best basis is:
{ ( 1, 0 ) , ( 0 , 1 ) }
)
So in this case the best basis would be:
1 0 0
0 0 0 = A1
0 0 0
0 1 0
0 0 0 = A2
0 0 0
0 0 1
0 0 0 = A3
0 0 0
0 0 0
0 1 0 = A4
0 0 0
0 0 0
0 0 1 = A5
0 0 0
0 0 0
0 0 0 = A6
0 0 1
to show that this set ( lets name it S ) of matrices is really basis for space of upper triangular matrices you need to show this:
if you have any upper triangular matrix:
x1 x2 x3
0 x4 x5 = X
0 0 x6
you can denote it like:
X = x1*A1 + x2*A2 + x3*A3 + x4*A4 + x5*A5 + x6*A6
so this is the proof that you can get any upper triangular matrix with linear combination of matrix from set S.
now you need to show that: ( y1, ... y6 ----> some variables )
0 0 0
0 0 0 = y1*A1 + y2*A2 + ... + y6*A6
0 0 0
is only on trivial way...
that is easy ( you need to show that this is true only if y1 = y2 = ... y6=0)
and that is the proof that S is basis for space of upper triangular matrices
c) iv
dont really know what you need to do ( dont understand ), but it cant be hard and if that is what i think it is than you will know what to do ( if you really know to solve problems c)i, ii, iii )
I agree that ci, cii, and ciii are much harder thatn a and b!
You are given the matrix $\displaystyle M= \begin{bmatrix}0 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 2\end{bmatrix}$ and your linear transformation is defined by T(A)= MA.
Any "vector" in V is a matrix of the form $\displaystyle \begin{bmatrix}a & b & c \\ 0 & d & e \\ 0 & 0 & f\end{bmatrix}$ which is
$\displaystyle \begin{bmatrix}0 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 2\end{bmatrix}\begin{bmatrix}a & b & c \\ 0 & d & e \\ 0 & 0 & f\end{bmatrix}= \begin{bmatrix}0 & 0 & 0 \\ 0 & d & e \\ 0 & 0 & 2f\end{bmatrix}$
ci) the range is the set of all possible results of that. Obviously (I hope it is obvious), that is the set of matrices of the form $\displaystyle \begin{bmatrix}0 & 0 & 0 \\ 0 & a & b \\ 0 & 0 & c\end{bmatrix}$ for some numbers a, b, c.
cii) the kernel is the set of all matrices A such that MA= 0 or $\displaystyle \begin{bmatrix}0 & 0 & 0 \\ 0 & d & e \\ 0 & 0 & 2f\end{bmatrix}= \begin{bmatrix}0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}$.
Setting comparable components of those equal, we must have d= 0, e= 0, and 2f= 0. a, b, and c can be anything so the kernel consists of matrices of the form $\displaystyle \begin{bmatrix}a & b & b \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{bmatrix}$.
ciii) The rank of T is the dimension of its image. That should be obvious: any of the three numbers a, b, c in $\displaystyle \begin{bmatrix}0 & 0 & 0 \\ 0 & a & b \\ 0 & 0 & c\end{bmatrix}$ can be chosen arbitrarily so the dimension of the image, and the rank of T is 3. (The kernel, in cii, also has dimension 3 so the "nullity" of T is 3, confirming the formula "rank(T)+ nullity(T)= dimension(V)".)
civ) Standard way of forming the matrix corresponding to a linear transformation, in a given basis: Apply the linear transformation to each of the basis vectors in turn. Write the result as a linear combination of the basis vectors. The coefficients form a column of the matrix.
Here: Apply T to each of the 6 basis vectors in (a). For example, the first basis "vector" josipive gave you was $\displaystyle A_1= \begin{bmatrix}1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{bmatrix}$. So $\displaystyle T(A_1)= \begin{bmatrix}0 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 2\end{bmatrix}\begin{bmatrix}1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{bmatrix}= \begin{bmatrix}0 & 0 & 0 \\0 & 0 & 0 \\ 0 & 0 & 0\end{bmatrix}$. (Here, $\displaystyle A_1$ is in the kernel of T) That can be written $\displaystyle 0A_1+ 0A_2+ 0A_3+ 0A_4+ 0A_5+ 0A_6$ so the first column in the matrix representation of T is a column of 6 zeros.