## AJBB one year ago Let {u,v,w} be a base for the vector space V. Determine if {u1,u2,u3} is a basis for V, where u1=u+v-3w; u2=u+3v-w and u3=v+w.

1. blockcolder

Given that any vector $$\vec{v}\in V$$ can be written as $$\vec{v}=c_1u+c_2v+c_3w$$, can $$\vec{v}$$ be also written as $$\vec{v}=d_1u_1+d_2u_2+d_3u_3$$?

2. ybarrap

$$Ax=b\\ \begin{bmatrix} 1&1 &-3 \\ 1&3 &-1 \\ 0& 1 & 1 \end{bmatrix} \begin{bmatrix} u\\ v\\ w \end{bmatrix} = \begin{bmatrix} u_1\\ u_2\\ u_3 \end{bmatrix}\\$$ Row-reducing A shows that $$u_1,u_2$$ and $$u_3$$ only fill a 2-dimensional space. $$\begin{bmatrix} 1&1 &-3 \\ 0&1 &1 \\ 0& 0 & 0 \end{bmatrix}$$ Since {u,v,w} is a basis for V, they are independent and fill a 3-dimensional space. Therefore, it is not possible for {u1,u2,u3} to be a basis for V.

3. AJBB

I have that the matrix is [1 1 0|X] [131|Y] [-3-11|] which I reduced and gave me 0=2x+1/2y+z I just don't know if it is correct.

4. ybarrap

I'm not following exactly. What i tried to show above was that the two basis could not span the same space because the second spans a smaller space than V.

5. AJBB

I have the matrix you showed, except that your rows are my columns and am trying to solve for a general vector (x,y,z) to what space it generates and then determine if it is a basis for V.

6. ybarrap

ok, I see that now.

7. Loser66

I know what you don't understand, you see, you said $${u_1,u_2,u_3}$$, it means you should expand them as column vectors. I mean |dw:1379732186029:dw|

8. AJBB

So I would have a matriz like this: |dw:1379732284595:dw|?

9. Loser66

yes

10. Loser66

and make it in the form of rref

11. ybarrap

That's right, now use @Loser66 's suggestion and row-reduce

12. Loser66

do you know how to make it in that form?

13. AJBB

Yes, I'm trying to do it right now.

14. Loser66

use your prof's terminology. Each of them has his own way to represent the process. And most of them ...hehehe.... tooooo stubborn to accept others' ways

15. Loser66

let see, It 's so late here. just post, if I can, I will.

16. AJBB

I get 0=2x+(1/2)y +z

17. Loser66

not that form, something like this |dw:1379733034406:dw|

18. Loser66

when you see 1 zero row in the system like this, it means the system is linear dependent.

19. Loser66

and if it is DEPENDENT system, it is not a base for R^3 which needs 3 INDEPENDENT vector base.

20. AJBB

Ok, got it! So I wouldn't need to find the determinant to see if it is a base?

21. Loser66

you can do it also. it's a good idea when you have a square matrix like this. det A =0, so it 's linear dependent--> not the base

22. Loser66

because it is general form to do for ANY matrix, that's why they give you this method. You are right when determine it by determinant

23. AJBB

ok, Thank you! Can you help me out with linear transformations, please?

24. Loser66

let see if I can,

25. AJBB

Let T:R3 to R2 be a linear transformation so that T(1,0,1)=(1,1) ; T(1,1,2)=(2,3) and T(1,1,0)=(2,1) Determine T(x,y,z) Prove that T(x,y,z) is a linear transformation. Sorry if the wording is a little weird, I'm translating form Spanish.

26. Loser66

I am not sure whether my answer is right or wrong because It 's easy to see. They give you that x, y ,z. Is it not that x =(1,0,1) ( first T), y =(1,1,2) ( second T ) and z = (1,1,0) ( the third T). And by calculate determinant of that matrix you have it = -2 therefore it 's linear independent .

27. AJBB

I don't know either. I think they are asking if it is a linear transformation not independent.I do know it has something to to with general vectors.

28. AJBB

|dw:1379734587983:dw|

29. Loser66

|dw:1379734719303:dw|