Question

NEED HELP ASAP!!
Let {u1, u2,u3} be a linearly independent set of vectors in R^n, and let vector v be a vector in R^n such that v= c1 u1+c2us+c3u3 for some scalars c1, c2,c3 wiht c3 not equal to 0. prove that the set {v, u1, u2} is linearly independent.

Answer 1

v= c1 u1+c2us+c3u3 is it c2 u2

Answer 2

argh I used to know this :*(

Answer 3

V=C1U1 +C2 U2 + C3U3

Answer 4

OK

Answer 5

Do you guys know how to do it?

Answer 6

Yes, Stay tuned

Answer 7

Thank you so much

Answer 8

Let @wio help you. My wife is telling me to go to dinner

Answer 9

Proof by contradiction might be a good way to go. If {v, u1, u2} is not linearly dependent, then {u1, u2,u3} is not.

Answer 10

If {v, u1, u2} is not linearly dependent, then we have constants \(k_1,k_2\) such that \[
v = k_1u_1+k_2u_2
\]

Answer 11

We know that \[
v = c_1 u_1+c_2u_2+c_3u_3 \implies \frac{1}{c_3}\left( v - c_1 u_1 -c_2u_2 \right)= u_3
\]

Answer 12

Now, substitute \(v=k_1u_1+k_2u_2\). Does it make sense?

Answer 13

I suppose this is actually a proof by contrapositive.

Answer 14

ok thank you :)

Answer 15

So you can take it from here? You know where I'm going?
Good!

Answer 16

Notice this proof wouldn't work if we didn't know beforehand that \(c_3\neq 0\).

Answer 17

@UsukiDoll I winged this one.

Answer 18

yay!

Answer 19

When I made the first post, I didn't actually know what I was going to do, and I just sort of came up with it as I went along.

Answer 20

XD

Answer 21

im sorry im actually stuck

Answer 22

No problem. Where are you stuck?

Answer 23

copy word for word XD

Answer 24

after substituting v= k1u1+K2u2

Answer 25

Okay, well, here is a hint. We want to show that \(u_3\) is linearly dependent on \(u_1,u_2\). This means that there are constants such that \[
u_3=d_1u_1+d_2u_2
\]You just need to find \(d_1,d_2\).

Answer 26

\[v=k_1u_1+k_2u_2\]

Answer 27

Do you understand how to find the \(d\)?

Answer 28

no thats where im confused

Answer 29

Okay so you have \[
u_3=\frac{1}{c_3}\left(k_1u_1+k_2u_2-c_1u_1-c_2u_2\right)
\]

Answer 30

combine like terms.

Answer 31

ok

Answer 32

what did you get?

Answer 33

(1/c3( k3u3-c1u1)

Answer 34

idk if its correct

Answer 35

Okay, so the problem is not just linear algebra, but also algebra.

Answer 36

yes

Answer 37

\[\begin{split}
u_3 &= \frac{1}{c_3}\left(k_1u_1+k_2u_2-c_1u_1-c_2u_2\right)\\
&= \frac{1}{c_3}\left((k_1-c_1)u_1+(k_2-c_2)u_2\right)\\
&=\left( \frac{k_1-c_1}{c_3}\right) u_1+\left(\frac{k_2-c_2}{c_3}\right)u_2\\
&=d_1 u_1+d_2u_2\\
\end{split}\]

Answer 38

aa i completely forgot how to combine it. it looks more familiar with numbers. im sorry.

Answer 39

now i have to find the values of d1 and d2

Answer 40

No, this completes the proof. We have shown that they exists, which means that \(u_1,u_2,u_3\) are not linearly independent.

Answer 41

ok I have to learn this topic more thoroughly. Doing an example helps. I have an exam tomorrow and I have no idea how to do this. Thank you for your help!! :D

Answer 42

I thought exams don't have proof problems D:

Answer 43

but the higher level courses do

Answer 44

unfortuanly my teacher is asking for proofs

Answer 45

w t fu u fuu

Answer 46

well if everyone scores bad, teacher's fault ^^

Answer 47

im not here to argue

Answer 48

o-0 no one is arguing

Answer 49

@anair , please follow me and I will prove it for you.

Answer 50

Thank you

Answer 51

yes thats correct.

Answer 52

v= c1 u1 + c2 u2 + c3 u3, suppose
a v + b u1 + d u2=0 then
a (c1 u1 + c2 u2 + c3 u3) +b u2 + d u2=0
(a c1 +b)u1 + (a c2+d )u2 + a c3 u3=0\\
a c1 + b =0
a c2 + d =0
a c3=0, since c3 is not zero, then a=0
a= 0 implies
0 c1 + b=0, then b =0
0 c2 + d=0 then d=0
We are done. Actually no need to consider case1 or case 2

Answer 53

a (c1 u1 + c2 u2 + c3 u3) +b u1 + d u2=0

Answer 54

There was a misprint in the third line bu2 should be b u1

Answer 55

Did you understand it?

Answer 56

yes i understand it much better.

Answer 57

Excellent

Answer 58

Sorry, I had to go to dinner, you cold have finished earlier.

Answer 59

Its ok thank you for ur help!

Answer 60

YW

Answer 61

:)

Answer 62

Is contrapositive too confusing?

Answer 63

yea it was