Question 378801: let T:R^3 --> R^3
T(x,y,z) = (z, x + y, x + y + z). determine
i) rank T and basis of Im T
ii) nullity T
what i tried:
If t is the matrix that will give you a 1 x 3 image then isn't is safe to conclude that the rank of T is 3? also if im T is the new vector, then the basis of im T is a linear combination of the standard basis vectors of R^3 right? but how do i prove all this mathematically?
Answer by robertb(5830)   (Show Source):
You can put this solution on YOUR website!  = (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  were the 1st column of the transformation matrix then  implies that a = b = 0, and c = 1.)
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