Question

Help please!

Find a basis of the subspace of R^4 spanned by the following vectors:

3 -1 3 -2
3 -4 3 1
-6 5 -6 1
0 -6 0 0

Answer 1

3 -1 3 -2 3 -4 3 1 -6 5 -6 1 0 -6 0 0 means ?

Answer 2

I don't know ;/

Answer 3

do I need to reduce to echelon form?

Answer 4

Seems like a plan. Let's see what you get.

Answer 5

there all individual matrices though?

Answer 6

like it's...

[ ] , [ ] , [ ] , [ ] and those numbers 4x1 in the brackets.

Answer 7

@satellite73 any idea as to how to solve this?

Answer 8

1st convert into rref then u will get the basis..

Answer 9

use gauss jordan to convert it into RREF

Answer 10

but they are individual matrices? not one full one...

Answer 11

Be that as it may, you can collect them into a matrix and manipulate them together.

They are not individual "matrices". They are individual "vectors". Why did you suggest RREF if you believed it inappropriate? Your instincts are good.

Answer 12

Sorry, vector, i used the wrong term. I didn't think you could do RREF of a 4x1 vector and I don't know how to collect them into a matrix and manipulate them together.. how do you do that?

Answer 13

yeah u have to take them as 1 matrix...4*4 matrix comprising of 4 vectors

Answer 14

ohh.. just put them together...? and then do RREF?

Answer 15

I got.....

1 0 1 -1
0 1 0 -1
0 0 0 0
0 0 0 0

Answer 16

so u have 2 free variables ...nd ur basis are (1,0,1,-1) & (0,1,0,-1)

Answer 17

It only gives me 67% when I insert...

1 0 x x
0 1 x x
1 0 x x
-1 -1 x x

Answer 18

I DIDN'T UNDERSTAND ?????????

Answer 19

All four are different vectors.. and if there are extra vectors you put x's in the boxes.

Answer 20

yup
nd then convert it into RREF

Answer 21

I did?

Answer 22

then?

Answer 23

I gave the answer in my previous comment ^

1 0 x x
0 1 x x
1 0 x x
-1 -1 x x

Answer 24

this is transpose of RREf ,this is not in RREF...

Answer 25

I'm so confused.. I already did rref once?

Answer 26

do I do it again?

Answer 27

chill,donot take transpose

Answer 28

no u did it correctly ,that matrix which u wrote was correct nd those answers which i wrote were its basis so ...question is over

Answer 29

but my program is only giving me 67% correct.. soo something is wrong..

Answer 30

ok,let me do it....wait

Answer 31

rref is 1 0 1 -2/3
0 1 0 0
0 0 0 0
0 0 0 0

Answer 32

still gives me 67%...

Answer 33

u r using just 2 basis na..????

Answer 34

and putting x's for all others yes.

Answer 35

no put 0

Answer 36

Which one is your first vector?

\([3,\;-1,\;3,\;-2]^{T}\)

or

\([3,\;3,\;-6,\;0]^{T}\)

Answer 37

just give 2 basis nd zeroes in place of 0...

Answer 38

^ first one

Answer 39

when I put 0's instead of x's it gives me 0%

Answer 40

omg

Answer 41

;/ idk what to do I've tried it 21 times.

Answer 42

What is this % thing? If you need a basis, you must find vectors.

Are we working the right problem? What sort of result are you expecting? What is the EXACT wording of the question? How are you trying to find a solution? Calculator?

I'm not quite sure we're speaking the same language.

Answer 43

the % is my score lol

Answer 44

The exact wording is up there ^..

Answer 45

I see. Well, it would help if we managed a correct RREF.

1) You have arranged the transposed vectors in a matrix form. This is perfect.

2) We need a proper Reduced Row Echelon Form.

3) Before we do that, notice that column 1 is identica to column 3. We ARE going to lose at least one vector when we get down to a basis. An R^3 subset is the best we will accomplish.

4) Reduced Row Echelon Form is \[\left[\begin{matrix}1& 0 & 1& 0\\ 0 & 1& 0& 0\\ 0& 0& 0& 1\\ 0& 0& 0& 0\end{matrix}\right]\]
I'm not sure where previous examples have been produced. You should redo your efforts until you manage this result.

5) Notice that we did, indeed, lose a vector. We didn't lose two of them.

6) If you need NORMAL basis vectors, you'll have to devide the first by \(\sqrt{2}\).

Answer 46

@mathhelp32 were you able to clear up the problem?

Answer 47

I'm a little confused still... so the RREF form of the matrix is that^?

Answer 48

@mathhelp32, that is correct, that is the RREF form of the matrix. So, your basis is going to include the first vector, the second vector, and the last vector.