Question

If X is the set consisting of the 6 vectors (1,1,0,0) (1,0,1,0) (1,0,0,1) (0,1,1,0) (0,0,1,1) (0,1,0,1) in C^4. Find 2 different maximal linearly independent subsets of X. (A maximal linearly independent subset of X is a linearly independent subset Y of X that becomes linearly dependent every time that a vector of X that is not already in Y is adjoined to Y) Please, help

Answer 1

@dan815

Answer 2

before I can even tell you if I have any idea how to answer this question, let me point out that you only have 5 vectors written when your problem states 6

Answer 3

oh, sorry, I missed one ( 0,1,0,1)

Answer 4

ok so I guess we need to come up with some linearly independent combinations of 4 of these 6 vectors, and check which ones will become linearly dependent if we tack on either of the other 2 we didn't use

Answer 5

go ahead, please.

Answer 6

I have never seen a problem like this either, so we're going to have to work together...
let's try some combinations
(1,1,0,0)
(1,0,1,0)
(1,0,0,1)
(0,1,1,0)
this set a linearly independent subset of X we can call Y, so that leaves the other members of X
(0,0,1,1) (0,1,0,1)
now if we "adjoin" one of these to Y, can it still be linearly independent?
I'm not sure what "adjoin" means here... do they mean replace?

Answer 7

Since we have 4 dimensions, so, the system has at most 4 linearly independent vectors. I can find it out by rref . However, I get just 1 set, it is {(1,1,0,0),(1,0,1,0),1,0,0,1),(0,1,1,0)} the 2left are linear combination of them.

Answer 8

maybe we are only supposed to select 3 linearly independent vectors for our subset Y, so we can adjoin another vector from Y' and not go outside C^4, which would make the set trivially linearly dependent

Answer 9

How about "maximal" term?

Answer 10

they define it in the problem as basically the most vectors of a set you can put together without getting linear dependency

Answer 11

Oooh, to satisfy this condition, the maximal linearly independent set much have only 3 vectors. Is it right?

Answer 12

That's how I see it, but again I'm trying to figure this out on the fly

Answer 13

and we can choose any 3 out of 4 linearly independent vectors above, right?

Answer 14

3/6 of the given vectors will make up Y, and it will be "maximal" if when we try to take any of the other 3 vectors we didn't put in Y and adjoin it and get linear dependency. If we start of with selecting 4 vectors, how can we talk about adjoining another?

Answer 15

so
(1,1,0,0)
(1,0,1,0)
(1,0,0,1)
is not maximal, because we can add (0,1,1,0) and it will still be linearly independent

Answer 16

no, no , I am wrong. Because the rest part said that when adding any other vector which are not in Y will turn Y be linearly dependent system. If we just pick 3 , when adding the left of the 4 , we still have linearly independent.

Answer 17

haaa!!! it's not easy!! how to do?? friend?

Answer 18

right now I'm at a trial-and-error level of approaching this problem :P

Answer 19

let's try Y=
(1,1,0,0)
(0,1,1,0)
(0,0,1,1)
is there another vector we can put on this set that leaves it linearly independent?

Answer 20

and yeah, we can write
(1,1,0,0)
(0,1,1,0)
(0,0,1,1)
(1,0,1,0) and that is also linearly dependent, so that doesn't work :(

Answer 21

I pick the first vector out (1,1,0,0) I have a system with a 4 linearly independent, too.

Answer 22

but we don't want a system of four LI vectors, we want a system of 3 LI vectors to which, if we add any one of the other vectors, makes the set LD

Answer 23

I mean I have 2 -4-linearly-independent systems,

Answer 24

ok, but how can we use that?

Answer 25

let say
1: (1,1,0,0)
2 : (1,0,1,0)
3(1,0,0,1)
4(0,1,1,0)
5(0,1,0,1)
6: (0,0,1,1)
the first system is (1,2,3,4)
the second one is (2,3,4,6)
they are LI
when adding any of the rest, I have LD

Answer 26

well of course, because you are trying to put 5 vectors in C^4
that's why I say I think we can only take them 3 at a time

Answer 27

3 vectors doesn't work as we said above, right?

Answer 28

I already showed that
(1,1,0,0)
(0,1,1,0)
(0,0,1,1)
are not maximal, so we can eliminate those

Answer 29

right

Answer 30

so now maybe try
(1,1,0,0)
(0,1,1,0)
(1,0,0,1)
then we have 3 other vectors to add

Answer 31

(1,1,0,0)
(0,1,1,0)
(1,0,0,1)
(0,0,1,1)
LI
so no good

Answer 32

darn seems like a lot of trial and error, there must be a smart way to do this
@phi

Answer 33

My prof asks us to do all problem on general way, not particular way. For example: when proving LI system of 2 vectors, I use det. He rejected. He said that what if I ask you to prove a system of 1000vectors, how can you take det? aaaaaah!!

Answer 34

lol, well I guess I'd use the method of cofactors, but I don't think there's a special fast way to get a determinant for a huge matrix like that. I wonder what he has in mind...

Answer 35

We are in Abstract Algebra course. Most students are behind the course, including me

Answer 36

Abstract? I thought it was just linear. Well, I'll keep pondering it, good luck in the meantime!

Answer 37

ty

Answer 38

linear is a subset of abstract:) but I would still call this linear.