Question 378801
{{{T(x,y,z) = (matrix(1,3,x,y,z))(matrix(3,3,0,1,1,0,1,1,1,0,1))}}} = (z, x + y, x + y + z).  The transformation matrix has rank 2, because column 3 is just the sum of columns 1 and 2.  Thus rank T = 2.  Now (z, x + y, x + y + z) = z(1,0,1) + (x+y)(0,1,1).  Hence {(1,0,1),(0,1,1)} is a basis for Im T.  Since rank T + nullity T = 3 (the number of columns of the transformation matrix), nullity T = 1.
(We found the transformation matrix by inspection.  Direct matrix multiplication would verify this. E.g., If {{{(matrix(3,1,a,b,c))}}} were the 1st column of the transformation matrix then {{{(matrix(1,3,x,y,z))*(matrix(3,1,a,b,c)) = z}}} implies that a = b = 0, and c = 1.)