Question

let vectors u,v,w be in R^n and suppose that the set (2u+w,u-v,w) is a basis for a subspace W. Show that the set (u,v,w) is a basis for W by showing that it satisfies the two conditions in the definition of a basis

Answer 1

{v1,v2,...,vn} is linearly independent
{v1,v2,...,vn} span W.

Answer 2

Got no clue right now. If there were numbers for (u,v,w) i'd be set..but this isn't the case.

Answer 3

anyone who can solve is god.

Answer 4

Why don't we proceed with the trickier part~ showing that u,v, and w are linearly indpendent?

Answer 5

*independent

Answer 6

Well clearly then the vectors 2u+w, u-v and w cannot be represented as a linear combination of each other as it forms a basis. So knowing this, how would we show that u,v,w are linearly independent?

Answer 7

I suggest we start with an assumption that a linear combination of u, v, and w is equal to the zero vector...
\[\Large a\color{red}{\vec u}+ b\color{green}{\vec v}+c\color{blue}{\vec w}= \vec0\]

Answer 8

Correct me if i'm wrong but then
a(2u+w)+b(u-v)+c(w) =0
implies that a=b=c=0?

Answer 9

Of course, you're right, but that equation has nothing to do with
\[\Large a\color{red}{\vec u}+ b\color{green}{\vec v}+c\color{blue}{\vec w}= \vec0\]

Answer 10

Now, let's rewrite this equation, but this time, in terms of the basis of W
2u + w
u - v
w

Answer 11

\[\Large a\color{red}{\vec u}+ b\color{green}{\vec v}+c\color{blue}{\vec w}=p(2\color{red}{\vec u}{+\color{blue}{\vec w}})+q(\color{red}{\vec u}-\color{green}{\vec v}) + r(\color{blue}{\vec w})= \vec0\]

Catch me so far?

Answer 12

Yep. Makes perfect sense.

Answer 13

Now, this will yield a system of equations... we just have to express p,q, and r
in terms of a,b, and c.

You can solve it as a systems, or you can use your incredible intuition to see a few 'details'...

Answer 14

*system

Answer 15

For instance, you'll see that the only element of the basis of W that involves v at all is the second one, (u - v)
And it has a -v in it, so you can bet that q must be -b.
\[\Large a\color{red}{\vec u}+ b\color{green}{\vec v}+c\color{blue}{\vec w}=p(2\color{red}{\vec u}{+\color{blue}{\vec w}})-b(\color{red}{\vec u}-\color{green}{\vec v}) + r(\color{blue}{\vec w})= \vec0\]

Answer 16

And now, you tentatively have a \(\large -b\color{red}{\vec u}\) involved, so you have to sort of "neutralise" that with the other element of the basis which involves u, the first one,
(2u + w)

2pu - bu = au
u(2p - b) = au
2p - b = a
2p = a+b

\[\large p = \frac{a+b}{2}\]

Answer 17

Yea i'm following. This is getting interesting now.

Answer 18

Well, stop following, and take the lead. ^.^

We now have
\[\Large a\color{red}{\vec u}+ b\color{green}{\vec v}+c\color{blue}{\vec w}=\left(\frac{a+b}{2}\right)(2\color{red}{\vec u}{+\color{blue}{\vec w}})-b(\color{red}{\vec u}-\color{green}{\vec v}) + r(\color{blue}{\vec w})= \vec0\]

Can you now find r (in terms of a,b, and c) ?

Answer 19

i'm doing that right now!

Answer 20

I'm waiting :)

Answer 21

is it \[c-(a+b)/2\]?

Answer 22

Very good :D
\[\Large a\color{red}{\vec u}+ b\color{green}{\vec v}+c\color{blue}{\vec w}=\left(\frac{a+b}{2}\right)(2\color{red}{\vec u}{+\color{blue}{\vec w}})-b(\color{red}{\vec u}-\color{green}{\vec v}) + \left[c-\frac{a+b}{2}\right](\color{blue}{\vec w})= \vec0\]

Answer 23

Crud.. let me fix that...
\[\large a\color{red}{\vec u}+ b\color{green}{\vec v}+c\color{blue}{\vec w}=\left(\frac{a+b}{2}\right)(2\color{red}{\vec u}{+\color{blue}{\vec w}})-b(\color{red}{\vec u}-\color{green}{\vec v}) + \left[c-\frac{a+b}{2}\right](\color{blue}{\vec w})= \vec0\]

Answer 24

Okay, that was the tricky part, the easy part comes now :D

Answer 25

YAY

Answer 26

Remember that our goal was to show that a,b, and c are all 0.
Right?

Answer 27

Yea...

Answer 28

Ok. would that mean that (a+b)/2=0,-b=0 and c-(a+b)/2=0?

Answer 29

Remembering as well that (2u + w) , (u-v) , and w
are linearly independent. That, combined with this equation...
\[\Large \left(\frac{a+b}{2}\right)(2\color{red}{\vec u}{+\color{blue}{\vec w}})-b(\color{red}{\vec u}-\color{green}{\vec v}) + \left[c-\frac{a+b}{2}\right](\color{blue}{\vec w})= \vec0\]

Is all we need to show that :)

and yes, you're correct... again :D

Answer 30

Oh!

Answer 31

So... solving this system...
\[\frac{a+b}2 = 0\]\[-b=0\]\[c-\frac{a+b}2=0\]

Answer 32

And don't skip it... I know you know what's going to happen, but solve anyway :P

Answer 33

a=-b,-b=0,2c=a+b

Answer 34

which means a=-b=0

Answer 35

and 2c=0
which implies c=0!?

Answer 36

as a=b=0

Answer 37

Okay... I'm satisfied. xD

So, since assuming au + bv + cw = 0
leads to
a = b = c = 0

We can conclude that u, v, and w are linearly independent :)

Answer 38

Nice Job!

Answer 39

We are not finished here, however.
We still have to show that
u, v, and w span W
In other words, any vector in W should be expressable as a linear combination of u, v, and w.


But I hope you can see that that's a ridiculously simple task... don't you think? ;)

Answer 40

would we then figure out that:

au+bw+cv=(x,y,z)?

or isn't it obvious that since these vectors are linearly independent to each other then any vector in W can be expressed as a linear combination of these vectors?

Answer 41

We've basically had enough of linear independence (we've shown all the linear independence needed, namely, that u, v, and w are independent)

Now, the task that remains is to show that any element of W can be written as a linear combination of u, v, and w.

Answer 42

Anyway, let's see... let
\[\large \nu \in W\]
then...
Since this set (2u + w) , (u - v), and w spans W, then...
\[\Large \nu = k(2\color{red}{\vec u}+\color{blue}{\vec w}) + m(\color{red}{\vec u }-\color{green}{\vec v}) + n\color{blue}{\vec w}\]
For some scalars k, m , and n.

Answer 43

Oh oops, that what i meant..for a moment i thought u,v,w were a basis.

Answer 44

So, can you continue from here?

Answer 45

It's just a matter of distributing the scalars, and separating the coefficients of u, v, and w.

Answer 46

let me have a go at that!

Answer 47

can i ask if this is right?

Answer 48

u(2k+m)+v(-m)+w(k+n)=v
Therefore
v=uX+vY+wZ, where X=2k+m, Y=-m, Z=k+n?

Answer 49

Hence this shows that (u,v,w) spans W?

Answer 50

I shouldn't have used v, maybe I should have used z instead :D
but yeah, you got it correctly :D
But I would place the scalars to the left of the vector, and not to the right, but that's a technicality ;)
Nicely done :D

Answer 51

So, recap, we showed that the set (u,v,w) is linearly independent, and also, that it spans W.

Therefore...?
^.^

Answer 52

You are the definition of a maths god! Thank you so much.

Answer 53

it satisfies the definition of a basis

Answer 54

That's the final bit, lol.

As for the "maths god"
hehe... I don't think so... ;)
If you're planning on sticking around, see if you can help others out with their woes :D
Welcome to Openstudy

I got to sign off now, so have fun ^.^

-------------------------------------------------
Terence out

Answer 55

Thanks mate!