Question

Let T : R3 to R3
be a linear transformation such that T (x1; x2; x3) =
(x1 + x2 + x3; x1 + 2x2 + 3x3; x1 + 3x2 + 5x3)

Answer 1

is T onto
is T a one to one?

Answer 2

i row reduced the matrix to
1 1 1
0 1 2
0 0 0

can someone please explain the answer

Answer 3

no and no

Answer 4

why is it not onto and one to one

Answer 5

what is T(1,-2,1) and T(2,-4,2)

Answer 6

T(1,-2,1) is (0, 0, 0)

Answer 7

yes

Answer 8

T(2,-4,2)=(0,0,0)

so T can not be 1-1

Answer 9

you have a nontrival null space. therefore it is not 1-1

Answer 10

your matrix only has rank 2 and therefore can only span a space of dimension 2. thus it is not onto

Answer 11

if your linear transformation can be represented by a square matrix and the matrix does not have full rank then it is not 1-1 and it is not onto

Answer 12

ok, and its not one to one because it has a free variable right? For the transformation to be one to one, there cant be free variables. Right?

Answer 13

yes...because it has a free variable its null space is nontrivial. Therefore it can't be 1-1