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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.

Linear Algebra
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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 \(\vec{v}=d_1u_1+d_2u_2+d_3u_3\)?
$$ Ax=b\\ \begin{bmatrix} 1&1 &-3 \\ 1&3 &-1 \\ 0& 1 & 1 \end{bmatrix} \begin{bmatrix} u\\ v\\ w \end{bmatrix} = \begin{bmatrix} u_1\\ u_2\\ u_3 \end{bmatrix}\\$$ 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.
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.

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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.
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.
ok, I see that now.
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|
So I would have a matriz like this: |dw:1379732284595:dw|?
yes
and make it in the form of rref
That's right, now use @Loser66 's suggestion and row-reduce
do you know how to make it in that form?
Yes, I'm trying to do it right now.
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
let see, It 's so late here. just post, if I can, I will.
I get 0=2x+(1/2)y +z
not that form, something like this |dw:1379733034406:dw|
when you see 1 zero row in the system like this, it means the system is linear dependent.
and if it is DEPENDENT system, it is not a base for R^3 which needs 3 INDEPENDENT vector base.
Ok, got it! So I wouldn't need to find the determinant to see if it is a base?
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
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
ok, Thank you! Can you help me out with linear transformations, please?
let see if I can,
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.
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 .
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.
|dw:1379734587983:dw|
|dw:1379734719303:dw|