anonymous
  • anonymous
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
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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blockcolder
  • 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 \(\vec{v}=d_1u_1+d_2u_2+d_3u_3\)?
ybarrap
  • 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.
anonymous
  • anonymous
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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ybarrap
  • 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.
anonymous
  • anonymous
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.
ybarrap
  • ybarrap
ok, I see that now.
Loser66
  • 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|
anonymous
  • anonymous
So I would have a matriz like this: |dw:1379732284595:dw|?
Loser66
  • Loser66
yes
Loser66
  • Loser66
and make it in the form of rref
ybarrap
  • ybarrap
That's right, now use @Loser66 's suggestion and row-reduce
Loser66
  • Loser66
do you know how to make it in that form?
anonymous
  • anonymous
Yes, I'm trying to do it right now.
Loser66
  • 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
Loser66
  • Loser66
let see, It 's so late here. just post, if I can, I will.
anonymous
  • anonymous
I get 0=2x+(1/2)y +z
Loser66
  • Loser66
not that form, something like this |dw:1379733034406:dw|
Loser66
  • Loser66
when you see 1 zero row in the system like this, it means the system is linear dependent.
Loser66
  • Loser66
and if it is DEPENDENT system, it is not a base for R^3 which needs 3 INDEPENDENT vector base.
anonymous
  • anonymous
Ok, got it! So I wouldn't need to find the determinant to see if it is a base?
Loser66
  • 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
Loser66
  • 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
anonymous
  • anonymous
ok, Thank you! Can you help me out with linear transformations, please?
Loser66
  • Loser66
let see if I can,
anonymous
  • anonymous
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.
Loser66
  • 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 .
anonymous
  • anonymous
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.
anonymous
  • anonymous
|dw:1379734587983:dw|
Loser66
  • Loser66
|dw:1379734719303:dw|