Basis

Printable View

May 21st 2009, 10:29 PM
Roam

Basis

Hi!

I don't understand how we can answer the following question.

How do we: extend the linearly independent set {(1,1,1,1,1),(1,2,3,4,5)} to form a basis for $R^5$ ?

Any help is appreciated.
May 21st 2009, 11:28 PM
scorpion007

Find 3 more linearly independent vectors to form a basis for $R^5$ . (You are given 2).

Recall that it is sufficient to find $n$ linearly independent vectors to form a basis for $R^n$ .
May 22nd 2009, 11:08 AM
Roam

Quote:

Originally Posted by scorpion007

Find 3 more linearly independent vectors to form a basis for $R^5$ . (You are given 2).

Recall that it is sufficient to find $n$ linearly independent vectors to form a basis for $R^n$ .

I'm not sure how to find thse vectors in $R^5$ ...
Would 'any' vector do? I mean can I use {(5,4,3,2,1),(6,7,8,9,1),(2,5,8,4,6)} for instance?
May 22nd 2009, 11:23 AM
Isomorphism

Quote:

Originally Posted by Roam

I'm not sure how to find thse vectors in $R^5$ ...
Would 'any' vector do? I mean can I use {(5,4,3,2,1),(6,7,8,9,1),(2,5,8,4,6)} for instance?

Sure, IF you can prove all the 5 together are linearly independent (Smirk)

Can you?
May 22nd 2009, 11:38 AM
SydneyBristow

Well the most simple basis to R^5
is the following 5 vectors because they are all linearly independant.

(1,0,0,0,0)
(0,1,0,0,0)
(0,0,1,0,0)
(0,0,0,1,0)
(0,0,0,0,1)

You have two vectors, you need 3 more that are linearly independant.

You can just easily use 3 of those or if you want to find more vectors you can u can just do trial and error.

The vectors you listed are probably linear combinations of the other vectors. You can check that by doing

v5= c1v1+c2v2+c3v3+c4v4

Usually if the vector contains a 0 component it is independant to a vector with numbers bc the c value would have to be 0 to make the vector 0 and if that happens it is unlikely to be a linear combination of the other vectors unless the other vectors have 0 in the same place.
May 22nd 2009, 07:41 PM
Roam

Okay, so can I just add the vectors {(0,0,1,0,0),(0,0,0,1,0),(0,0,0,0,1)} to the set in order to extend the linearly independent set {(1,1,1,1,1),(1,2,3,4,5)} to form a basis for $R^5$ ?

To find out whether these vectors are linearly independent together I must test them. So, for scalars $c_{1},c_{2},c_{3},c_{4},c_{5} \in R$ , in the linear combination:

$c_{1} (1,1,1,1,1) + c_{2}(1,2,3,4,5) + c_{3} (0,0,1,0,0) + c_{4}(0,0,0,1,0) + c_{5}(0,0,0,0,1)$

There can only be one solution to the scalars, $c_{1},c_{2},c_{3},c_{4},c_{5} = 0$

I know the answer to this question isn't unique but is my answer correct? Is this a right set (and a basis for $R^5$ )?
May 22nd 2009, 08:56 PM
scorpion007

Yes. (You can verify it by solving the system you mentioned)