Linear transformation being onto or not (Three problems I need help understanding)

  • Thread starter sam0617
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  • #1
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Which of the transformations are onto?

1) T:R2 -> R2, where T(x,y) = (5x-y, 0)

I don't know if I'm understanding this correctly but this transformation is NOT onto because if I let
5x-y = a
0 = b

this means that b doesn't cover all the range of T? Could someone explain it better if I'm wrong.


2) T:R3 -> R2 where T(x,y,z) = (x+y, x-z)
so I equate this to
x+y = a
x-z = b
which makes the matrix
row1 = (1 1 0 a)
row 2 = (1 0 -1 b)
then once I reduce it to row echelon form, I ultimately get
row 1 = (1 0 -1 b)
row 2 =( 0 1 1 a-b)

then I get stuck because I don't understand what that means..I'm going to guess it's NOT onto because for any value a and b, I can't get and x,y, or z?

Please correct me if I'm wrong.


3) T:R2 -> R3, where T(x,y) = (y, x, x-y)
I equate this again to
y = a
x= b
x-y = c
which forms the matrix
row 1 = (0 1 a)
row 2 =(1 0 b)
row 3 = (1 -1 c)
then perform row echelon to ultimately get
row 1 = (1 0 b)
row 2 = (0 1 a)
row 3 = (0 -1 -b+c)
I think this is NOT onto again because of the last row but I can't be 100% sure. Any guidance would help a lot.

I know this is much to ask but if you can even help me with just one, it would mean a lot. Thank you for any help.
 

Answers and Replies

  • #2
Deveno
Science Advisor
906
6


there's different ways to approach these problems.

1) your reasoning here is correct, because the second coordinate of T(x,y) is always 0, so if we have T(x,y) = (a,b), and b ≠ 0, then T(x,y) = (a,b) has no solution. for example, there is no (x,y) such that T(x,y) = (0,1).

2) we can use the same approach as (1):

x+y = a
x-z = b

from the first equation, y = a-x.
from the second, z = x-b, so (x,a-x,x-b) = x(1,-1,1) + (0,a,-b) should be a solution, for any x. we only need one, so let's choose x = 0.

T(0,a,-b) = (0+a,0-(-b)) = (a,b), no matter what a and b are. we have specifically exhibited a pre-image (in fact, a lot of them, depending on what we choose for x) for (a,b), so T is onto.

if we want to use matrices, we need to write T in terms of some basis. the basis {(1,0,0),(0,1,0), (0,0,1)} works pretty well. the first column for T is T(1,0,0) = (1,1). the second column is T(0,1,0) = (1,0), and the third column is T(0,0,1) = (0,-1). so the augmented matrix is A' =

[tex]\begin{bmatrix}1&1&0&a\\1&0&-1&b \end{bmatrix}[/tex]

as you correctly stated. however, this reduces to:

[tex]\begin{bmatrix}1&0&-1&b\\0&1&1&a-b \end{bmatrix}[/tex]

what does this mean? it means that any solution (x,y,z) to T(x,y,z) = (a,b),

also satisfies:

[tex]\begin{bmatrix}1&0&-1\\0&1&1\end{bmatrix} \begin{bmatrix}x\\y\\z \end{bmatrix} = \begin{bmatrix}b\\a-b \end{bmatrix}(*)[/tex]

that is: x-z = b, and y+z = a-b. since we only have 2 rows, we can eliminate z from our equations (say, assigning it the value t), so that we get:

x = b+z = b+t
y = a-b-t, that is: (x,y,z) = (b+t,a-b-t,t) = t(1,-1,1) + (b,a-b,0).

and, indeed T(t(-1,1,1) + (b,a-b,0)) = tT(1,-1,1) + T(b,a-b,0) = t(0,0) + (b+a-b,b-0) = (a,b).

but (*) tells us more. it tells us that rank(T) = 2, and since dim(R2) = 2 = rank(T) = dim(im(T)), im(T) is a 2-dimensional subspace of R2, that is, im(T) is ALL of R2, so T is onto (we don't even have to actually calculate the solutions to see this).

we also see that the nullspace of A = ker(T) is dependent only on the single parameter t, and that it has the basis {(1,-1,1)} (this is the solution when a = b = 0), so dim(ker(T)) = nullity(A) = 1. this illustrates a general principal of linear algebra:

general solution = homogeneous solutions + specific solution.

note that our two methods give the same basis for ker(T), but two different "specific" solutions:

(b,a-b,0) and (0,a,-b). but these are both examples of the general solution:

t(1,-1,1) + (b,a-b,0). for t = 0, we get (b,a-b,0). for t = -b, we get (-b,b,-b) + (b,a-b,0) = (0,a,-b).

look at problem 3 again, in light of this.
 

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