Got Homework?
Connect with other students for help. It's a free community.
Here's the question you clicked on:
 0 viewing
AJBB
Group Title
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+v3w; u2=u+3vw and u3=v+w.
 10 months ago
 10 months ago
AJBB Group Title
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+v3w; u2=u+3vw and u3=v+w.
 10 months ago
 10 months ago

This Question is Closed

blockcolder Group TitleBest ResponseYou've already chosen the best response.0
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 \(\vec{v}=d_1u_1+d_2u_2+d_3u_3\)?
 10 months ago

ybarrap Group TitleBest ResponseYou've already chosen the best response.0
$$ Ax=b\\ \begin{bmatrix} 1&1 &3 \\ 1&3 &1 \\ 0& 1 & 1 \end{bmatrix} \begin{bmatrix} u\\ v\\ w \end{bmatrix} = \begin{bmatrix} u_1\\ u_2\\ u_3 \end{bmatrix}\\$$ Rowreducing A shows that \(u_1,u_2\) and \(u_3\) only fill a 2dimensional 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 3dimensional space. Therefore, it is not possible for {u1,u2,u3} to be a basis for V.
 10 months ago

AJBB Group TitleBest ResponseYou've already chosen the best response.0
I have that the matrix is [1 1 0X] [131Y] [311] which I reduced and gave me 0=2x+1/2y+z I just don't know if it is correct.
 10 months ago

ybarrap Group TitleBest ResponseYou've already chosen the best response.0
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.
 10 months ago

AJBB Group TitleBest ResponseYou've already chosen the best response.0
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.
 10 months ago

ybarrap Group TitleBest ResponseYou've already chosen the best response.0
ok, I see that now.
 10 months ago

Loser66 Group TitleBest ResponseYou've already chosen the best response.0
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
 10 months ago

AJBB Group TitleBest ResponseYou've already chosen the best response.0
So I would have a matriz like this: dw:1379732284595:dw?
 10 months ago

Loser66 Group TitleBest ResponseYou've already chosen the best response.0
and make it in the form of rref
 10 months ago

ybarrap Group TitleBest ResponseYou've already chosen the best response.0
That's right, now use @Loser66 's suggestion and rowreduce
 10 months ago

Loser66 Group TitleBest ResponseYou've already chosen the best response.0
do you know how to make it in that form?
 10 months ago

AJBB Group TitleBest ResponseYou've already chosen the best response.0
Yes, I'm trying to do it right now.
 10 months ago

Loser66 Group TitleBest ResponseYou've already chosen the best response.0
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
 10 months ago

Loser66 Group TitleBest ResponseYou've already chosen the best response.0
let see, It 's so late here. just post, if I can, I will.
 10 months ago

AJBB Group TitleBest ResponseYou've already chosen the best response.0
I get 0=2x+(1/2)y +z
 10 months ago

Loser66 Group TitleBest ResponseYou've already chosen the best response.0
not that form, something like this dw:1379733034406:dw
 10 months ago

Loser66 Group TitleBest ResponseYou've already chosen the best response.0
when you see 1 zero row in the system like this, it means the system is linear dependent.
 10 months ago

Loser66 Group TitleBest ResponseYou've already chosen the best response.0
and if it is DEPENDENT system, it is not a base for R^3 which needs 3 INDEPENDENT vector base.
 10 months ago

AJBB Group TitleBest ResponseYou've already chosen the best response.0
Ok, got it! So I wouldn't need to find the determinant to see if it is a base?
 10 months ago

Loser66 Group TitleBest ResponseYou've already chosen the best response.0
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
 10 months ago

Loser66 Group TitleBest ResponseYou've already chosen the best response.0
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
 10 months ago

AJBB Group TitleBest ResponseYou've already chosen the best response.0
ok, Thank you! Can you help me out with linear transformations, please?
 10 months ago

Loser66 Group TitleBest ResponseYou've already chosen the best response.0
let see if I can,
 10 months ago

AJBB Group TitleBest ResponseYou've already chosen the best response.0
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.
 10 months ago

Loser66 Group TitleBest ResponseYou've already chosen the best response.0
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 .
 10 months ago

AJBB Group TitleBest ResponseYou've already chosen the best response.0
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.
 10 months ago

AJBB Group TitleBest ResponseYou've already chosen the best response.0
dw:1379734587983:dw
 10 months ago

Loser66 Group TitleBest ResponseYou've already chosen the best response.0
dw:1379734719303:dw
 10 months ago