## AJBB 2 years ago T:R3 -> R2 is a linear transformation where T(1,1,2)=(2,3) T(1,0,1)=(1,1) and T(1,1,0)=(2,1) Determine T(X,Y,Z) and prove that T(X,Y,Z) is a linear transformation.

1. blockcolder

$A\begin{bmatrix}1\\1\\2\end{bmatrix}=\begin{bmatrix}2\\3\end{bmatrix}; A\begin{bmatrix}1\\0\\1\end{bmatrix}=\begin{bmatrix}1\\1\end{bmatrix}; A\begin{bmatrix}1\\1\\0\end{bmatrix}=\begin{bmatrix}2\\1\end{bmatrix}$ If you let $A=\begin{bmatrix}a_{11}&a_{12}&a_{13}\\ a_{21}&a_{22}&a_{23}\\ \end{bmatrix}$ Then you will get two system of equations so that you can solve for the entries of A.

2. AJBB

I don't quite understand. Could you please explain?

3. Zarkon

put the following matrix into rref $\left[\begin{array}{ccccc} 1 & 1 & 2 & 2 & 3 \\ 1 & 0 & 1 & 1 & 1 \\ 1 & 1 & 0 & 2 & 1 \\ \end{array}\right]$

4. AJBB

On it.

5. AJBB

Could you tell me what for, please?

6. Zarkon

it is a way of finding the answer...it will give you what T(1,0,0) T(0,1,0) and T(0,0,1) are

7. Zarkon

then T(x,y,z)=xT(1,0,0)+yT(0,1,0)+zT(0,0,1)

8. AJBB

Is this correct |dw:1379741449340:dw| ?

9. Zarkon

no

10. Zarkon

$\left[\begin{array}{ccccc} 1 & 0& 0 & 1 & 0 \\ 0 & 1 & 0 & 1 & 1 \\ 0 & 0 & 1 & 0 & 1 \\ \end{array}\right]$

11. AJBB

Ok. And then what should I do?

12. Zarkon

then T(1,0,0)=(1,0) T(0,1,0)=(1,1) T(0,0,1)=(0,1)

13. Zarkon

T(x,y,z)=xT(1,0,0)+yT(0,1,0)+zT(0,0,1) =x(1,0)+y(1,1)+z(0,1)=(x+y,y+z)

14. AJBB

Could you explain this with words please? I am not following you.

15. Zarkon

I just fond out where the standard basis vectors get mapped to. Once you have those the problem is easy

16. Zarkon

if this is confusing you can always try blockcolder's method...that will work too.

17. AJBB

I just don't understand this at all. I want too, I just can't seem to get it in my head.

18. Zarkon

do you understand where my original matrix came from?

19. AJBB

No, not really. I would really appreciate it if you could explain it step-by-step, please.

20. Zarkon

look at $\left[\begin{array}{ccc|cc} 1 & 1 & 2 & 2 & 3 \\ 1 & 0 & 1 & 1 & 1 \\ 1 & 1 & 0 & 2 & 1 \\ \end{array}\right]$

21. Zarkon

look at the first row...it is just T(1,1,2)=(2,3) 1 1 2 | 2 3

22. AJBB

Ok.

23. Zarkon

same with the 2nd and 3rd rows

24. Zarkon

then you row reduce it....this gives us what the standard basis vectors map to

25. AJBB

What is a standard basis map and what is it for? I'm sorry for all my questions. I just saw this yesterday in class and my test is tomorrow.

26. Zarkon

we find the standard basis...(1,0,0), (0,1,0) and (0,0,1) and what their image is under T is ie the values of T(1,0,0),T(0,1,0) and T(0,0,1)

27. AJBB

And then what do we do?

28. Zarkon

the wording i'm using mean the following T(1,1,2)=(2,3) is the same as saying T maps (1,1,2) to (2,3)

29. Zarkon

then we use the linearity of T to get the answer

30. Zarkon

we want T(x,y,z)

31. Zarkon

T[(x,y,z)]=T[(x,0,0)+(0,y,0)+(0,0,z)] =T(x,0,0)+T(0,y,0)+T(0,0,z) =xT(1,0,0)+yT(0,1,0)+zT(0,0,1)

32. AJBB

and then?

33. Zarkon

T[(x,y,z)]=T[(x,0,0)+(0,y,0)+(0,0,z)] =T(x,0,0)+T(0,y,0)+T(0,0,z) =xT(1,0,0)+yT(0,1,0)+zT(0,0,1) =x(1,0)+y(1,1)+z(0,1) =(x,0)+(y,y)+(0,z) =(x+y,y+z)

34. Zarkon

so T(x,y,z)=(x+y,y+z) if we plug in our original vectors T(1,1,2)=(1+1,1+2)=(2,3) T(1,0,1)=(1+0,0+1)=(1,1) T(1,1,0)=(1+1,1+0)=(2,1) so it looks like we are correct

35. AJBB

but where do you get (x,0),(y,y) and (0,z)?

36. Zarkon

from our rref matrix $\left[\begin{array}{ccc|cc} 1 & 0& 0 & 1 & 0 \\ 0 & 1 & 0 & 1 & 1 \\ 0 & 0 & 1 & 0 & 1 \\ \end{array}\right]$

37. Zarkon

T(1,0,0)=(1,0) first row of matrix 1 0 0 | 1 0

38. AJBB

oh, got it!

39. Zarkon

good

40. AJBB

Would it be the same process for linear transformations that for up instead of down,s ay from R2 -> R3?

41. Zarkon

yes

42. AJBB

Ok! By any chance, do you have practice problems or do you know where I could find any?

43. Zarkon

none typed up

44. Zarkon

you have a text book?

45. AJBB

I do, but it doesn't have linear transformations in it.

46. Zarkon

what book do you have?

47. AJBB

Found them! I was looking in the wrong chapter. Anyways, its Linear Algebra by Bernard Kolman and David R. Hill.

48. Zarkon

ok...I have a lot of linear algebra book...but I don't think I have that one

49. Zarkon

*books

50. Zarkon

I have like 10 of them...i'd have to check my office ;)

51. AJBB

Its ok. My book is in Spanish anyway. How much longer are you going to stay around?

52. Zarkon

not much longer...it is pretty late for me

53. Zarkon

2am

54. AJBB

Ok! Thank you so much for your help, patience and time!

55. Zarkon

no problem...good luck on your exam!

56. AJBB

Thank you!