Question

Find 2 bases in C^4such that the only vectors common to both are (0,0,1,1) and (1,1,0,0)
Please help me put it in neat.

Answer 1

(0,0,1,1)+(1,1,0,0)= (1,1,1,1)
therefore the vector sum must be not (1,1,1,1)
choose
a) (0,1,1,1)
b) (1,0,1,1)
c) (1,1,1,0)
d) (1,1,0,1)
pick 2 of them combine with the given vectors to form the bases. The answer are
{(0,0,1,1), (1,1,0,0),(0,1,1,1),(1,0,1,1)} and {(0,0,1,1), (1,1,0,0),(1,1,1,0),(1,1,0,1)}

Answer 2

@ganeshie8

Answer 3

can u define the bases plz ?

Answer 4

I got them, just don't know how to put the logic in algebraic way.

Answer 5

huh , define it cuz i wanna know wat it is ..

Answer 6

{(0,0,1,1), (1,1,0,0),(0,1,1,1),(1,0,1,1)}
{(0,0,1,1), (1,1,0,0),(1,1,1,0),(1,1,0,1)}

Answer 7

hmm u dnt got me
for any 2 vectors
what is the definition of bases ?

Answer 8

linearly independent

Answer 9

@wio

Answer 10

Hmmm, interesting question

Answer 11

So would (0,0,2,2) also be in both bases, or does magnitude also matter here?

Answer 12

since it is a base, so I think it should be unit one.

Answer 13

Well, do you know any particular method to do this?
I know that you can use rref to show that the bases has linearly independent vectors.

Answer 14

We have 2given vectors which are linearly independent. So , need 4 more to have 2 bases

Answer 15

I got them, but you mean I have to prove that they are linearly independent, right?

Answer 16

If it is, just find det

Answer 17

The easiest way to verify that your bases are independent after you've picked which vectors go to which bases is by showing linear independence.

Answer 18

You are in complex space. The basis could include imaginary elements.

Answer 19

hey, so you assume that 0 in (1,1,0.0) are imaginary part?

Answer 20

C^4 is in complex 4-space, so I'm thinking you can take advantage of that

Answer 21

What about
( i , -i , 1 ,1)
(1 , 1 , i, -i )

With your original
(1, 1, 0, 0)
(0, 0, 1, 1)

They are all orthogonal and contain the latter 2 vectors.