## rsst123 one year ago Let T:R3->R3 be the linear transformation such that T(x1,x2,x3) = (x1+2x2+2x3,x1+2x2+x3,x1-2x2+2x3) a) is T 1-1? Briefly explain b) is T onto? Briefly explain I know T is 1-1 if the only solution to T(x)=0 is x=0 and for onto I'm not really sure. Can someone please help me understand 1-1 and onto and how to solve this problem. Thanks!

1. dan815

|dw:1434823327570:dw|

2. dan815

ya

3. ParthKohli

So first of all, we have to really refer to the definitions. What does a one-one transformation really mean? It tells you if there's only one way to get to a point after a transformation. You can't start with two same points and end up with the same point after the transformation. Let's assume that there *are* two such points, say, $$(x_1, x_2, x_3)$$ and $$(y_1, y_2, y_3)$$. Now $$T(x_1, x_2, x_3) = (x_1+2x_2+2x_3,~x_1+2x_2+x_3,~x_1-2x_2+2x_3)$$ and $$T(y_1, y_2, y_3) = (y_1 + 2y_2 + 2y_3, ~ y_1 + 2y_2 + y_3, ~y_1 - 2y_2 + 2y_3)$$. If these two points are identical for unidentical triples $$(x_1, x_2, x_3); (y_1, y_2, y_3)$$ then we can claim that the function is not one-one. This means:$x_1+2x_2+2x_3 =y_1 + 2y_2 + 2y_3$$x_1+2x_2+x_3 =y_1 + 2y_2 + y_3$$x_1-2x_2+2x_3 = y_1 - 2y_2 + 2y_3$

4. ParthKohli

Let's write these equations as:$\lambda_1 + 2\lambda_2 + 2\lambda_3 = 0$$\lambda_1 + 2\lambda_2 + \lambda_3 = 0$$\lambda_1 - 2 \lambda_3 + 2\lambda_3=0$Where $$\lambda_1 = x_1 - y_1$$ and so on. Turns out that the only solution here is the trivial one: $$\lambda_1 = \lambda_2 = \lambda_3 = 0$$.

5. ParthKohli

Therefore, $$x_1 = y_1$$; $$x_2 = y_2$$ and $$x_3 = y_3$$ making this a one-one function.

6. ParthKohli

What I meant in my first post was "you can't start with two different points and end up with the same point".

ok so the points for x and y have to be the same to be 1-1 right? and is onto just the opposite when two different points and end up with the same point?

8. dan815

the matrix is linearly independant

9. ParthKohli

What does an onto-transformation mean? It means that the transformation covers all points. If I take any point, say (1, 1, 2), then I can find a point that transforms to give that point (here, (1, 1, 2)).

10. ParthKohli

So let's say I have a point $$(x,y, z)$$ then I have to show that there exists a point $$(k_1, k_2, k_3)$$ such that $$T(k_1, k_2, k_3) = (x,y,z)$$ if this is an onto function.

11. ParthKohli

Assume that $$T(x_1, x_2, x_3) = (x,y,z)$$ for some $$(x_1, x_2, x_3)$$. By the definition of our transformation, this point $$(x,y,z)$$ is the same as $$(x_1+2x_2+2x_3,x_1+2x_2+x_3,x_1-2x_2+2x_3)$$. Therefore,$x_1 + 2x_2 + 2x_3 = x$$x_1 + 2x_2 + x_3 = y$$x_1 - 2x_2 + 2x_3 = z$Sounds good so far. What do we really have to do here? We have to express $$x_1, x_2,x_3$$ in terms of $$x, y, z$$ to show that this function is onto. Can we do that? Yup, we can.

12. ParthKohli

For instance, $$x_3 = x - y$$ from the first two equations.

13. ParthKohli

Similarly, $$x_1 = \dfrac{y+z -3x_3}2 = \dfrac{y + z - 3x + 3y}{2} =\dfrac{4y - 3x + z}{2}$$

14. ParthKohli

So basically what I did here was that I actually found the point that maps to $$(x,y,z)$$ to show that this is an onto function.

15. ParthKohli

If you know basic linear algebra, Dan's solution is actually WAYYYY simpler than mine is.

ok so a simple answer would be if its linearly independent its onto and if not 1-1?