Linear Algebra help :(

[BamBam215]

I'm given a vector space V=u1, u2, u3, u4, u5, u6, u7, u8. From the list of 8 vectors, I need to find the ones that form a basis for V, in other words I have to find the ones that are linearly independent.

So say for example, I line up u1, u2, u3, u4 into matrix form and perform elementary row operations. I get the following results:

[ 1 3 2 5 1 ] <--- u1
[ 0 1 3 2 1 ] <--- u2
[ 0 0 0 0 0 ] <--- u3
[ 0 0 0 0 0 ] <--- u4

So is it safe to say that u3 and u4 are linearly dependent or is it the other way around (u1 and u2 are linearly dependent)??? This seems very easy but I lost my train of thought and it's bugging me

 

[jinduy]

nightmares

 

[Mday]

Originally posted by: BamBam215
I'm given a vector space V=u1, u2, u3, u4, u5, u6, u7, u8. From the list of 8 vectors, I need to find the ones that form a basis for V, in other words I have to find the ones that are linearly independent.

So say for example, I line up u1, u2, u3, u4 into matrix form and perform elementary row operations. I get the following results:

[ 1 3 2 5 1 ] <--- u1
[ 0 1 3 2 1 ] <--- u2
[ 0 0 0 0 0 ] <--- u3
[ 0 0 0 0 0 ] <--- u4

So is it safe to say that u3 and u4 are linearly dependent or is it the other way around (u1 and u2 are linearly dependent)??? This seems very easy but I lost my train of thought and it's bugging me

anyway, that 4x5 matrix tells you this: u3 and u4 are linear combinations of u1 and u2. you can manipulate them to say that u1 and u3 are linear combinations of u2 and u4. with that example you gave, you can say that u1 and u2 form a basis for your vector space comprised of u1, u2 and u3.

if you are very well versed in what it means to be linearly dependent, you would be able to figure more things out. so, look back to what it really means... and recall what a basis is. and recall that a basis is NOT unique.

--

if you are given 8 vectors, and each has 5 entries, you must realize that the basis is at MOST 5. if it's less than 5, it tells you the true dimension of the vector space, which is not always the same as the number of entries in a vector. oh, if you have more than 5, you made errors.

 

[BamBam215]

Originally posted by: Mday

Originally posted by: BamBam215
I'm given a vector space V=u1, u2, u3, u4, u5, u6, u7, u8. From the list of 8 vectors, I need to find the ones that form a basis for V, in other words I have to find the ones that are linearly independent.

So say for example, I line up u1, u2, u3, u4 into matrix form and perform elementary row operations. I get the following results:

[ 1 3 2 5 1 ] <--- u1
[ 0 1 3 2 1 ] <--- u2
[ 0 0 0 0 0 ] <--- u3
[ 0 0 0 0 0 ] <--- u4

So is it safe to say that u3 and u4 are linearly dependent or is it the other way around (u1 and u2 are linearly dependent)??? This seems very easy but I lost my train of thought and it's bugging me

anyway, that 4x5 matrix tells you this: u3 and u4 are linear combinations of u1 and u2. you can manipulate them to say that u1 and u3 are linear combinations of u2 and u4. with that example you gave, you can say that u1 and u2 form a basis for your vector space comprised of u1, u2 and u3.

if you are very well versed in what it means to be linearly dependent, you would be able to figure more things out. so, look back to what it really means... and recall what a basis is. and recall that a basis is NOT unique.

So in other words u1 and u2 are linearly independent and are two of the vectors that form a basis for V?

That was my initial assumption but then I ran into something weird. I found that u1, u3, u5 are three vectors in the basis. Then I tested those 3 vectors with u6 and found the following:

[ 1 3 2 2 1 ] <-u1
[ 0 1 6 1 7 ] <-u3
[ 0 0 0 0 0 ] <-u5
[ 0 0 0 0 0 ] <-u6


So now u5 is linearly dependent even though I found it to be linearly indepedent before?

 

[Mday]

problem, is, you didnt do all 8 at once. reduce it to whatever.

recall the initial sample matrix with u1, 2 3 and 4.

i told you the way you can state is have u1 and u2 be the basis of that specific matrix (not the V in your problem). OR u1 and u3 could be. or even u2 and u3, or u4 and u1. a basis is not unique when an entire "row" becomes 0.

 

[BamBam215]

aye so i should row reduce the entire 8 vectors at once? the book does it one by one.

 

[Mday]

Originally posted by: BamBam215
aye so i should row reduce the entire 8 vectors at once? the book does it one by one.

doing it one by one tells you one thing: if two vectors are linearly independent. if they are linearly dependent, you just reduced the possibilities for ONE BASIS. if they are linearly independent, you pretty much have 8 choose 2, which is huge, more pairs to go.

anyway.

you should reduce one row at a time. with the full 8 vectors in a matrix. if you keep doing 4 vectors at a time, you will run into the same mess you have now.