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 and your linear transformation is defined by T(A)= MA.
Any "vector" in V is a matrix of the form which is
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 for some numbers a, b, c.
cii) the kernel is the set of all matrices A such that MA= 0 or .
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 .
ciii) The rank of T is the dimension of its image. That should be obvious: any of the three numbers a, b, c in 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 . So . (Here, is in the kernel of T) That can be written so the first column in the matrix representation of T is a column of 6 zeros.