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.

- anonymous

- jamiebookeater

See more answers at brainly.com

At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga.
Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus.
Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this

and **thousands** of other questions

- 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

$$
Ax=b\\
\begin{bmatrix}
1&1 &-3 \\
1&3 &-1 \\
0& 1 & 1
\end{bmatrix}
\begin{bmatrix}
u\\
v\\
w
\end{bmatrix}
=
\begin{bmatrix}
u_1\\
u_2\\
u_3
\end{bmatrix}\\$$
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

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.

Looking for something else?

Not the answer you are looking for? Search for more explanations.

## More answers

- 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

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

ok, I see that now.

- 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

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

- Loser66

yes

- Loser66

and make it in the form of rref

- ybarrap

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

- Loser66

do you know how to make it in that form?

- anonymous

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

- 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

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

- anonymous

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

- Loser66

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

- Loser66

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

- Loser66

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

- anonymous

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

- 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

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

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

- Loser66

let see if I can,

- 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

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

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

|dw:1379734587983:dw|

- Loser66

|dw:1379734719303:dw|