Lin Alg - Rank of a Matrix

1. Jul 20, 2010

jinksys

Let S = {v1, v2, v3, v4, v5}

v1 = <1,1,2,1>
v2 = <1,0,-3,1>
v3 = <0,1,1,2>
v4 = <0,0,1,1>
v5 = <1,0,0,1>

Find a basis for the subspace V = span S of R^4.

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My attempt:

I place the five vectors into a matrix, where each vector is a row of the matrix.
I solve for row-echelon (not RREF). I get:

Code (Text):

1 1 2 1
0 1 5 0
0 0 1 0
0 0 0 1
0 0 0 0

Therefore the basis of V = span S of R4 contains the vectors:

<1,1,2,1>, <0,1,5,0>, <0,0,1,0>, <0,0,0,1>

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Question, besides the question "Is this correct?", I'd like to know the difference between solving for REF and RREF. I have two textbooks that I'm using to get through LinAlg and they differ on this section. One tells me to solve for RREF and one REF. Using the former method I get the identity matrix for this problem, and using the latter I get the basis above.

Last edited: Jul 20, 2010
2. Jul 20, 2010

hunt_mat

Not too sure what this is asking. span(S)=R^4 (ignore v1 and you can find yourself an orthonormal basis). Your working is correct, why did you leave out the fifth vector?

Mat

3. Jul 20, 2010

jinksys

Good catch, left out a row of zeros.

Are you not sure what I am asking, or the problem?

4. Jul 20, 2010

hunt_mat

Both, S clearly spans R^4, so one of the vectors must be a linear combination of the others.
Can you define what is meant by REF and RREF(row reduced echlon form?)

Mat

5. Jul 20, 2010

jinksys

REF = row echelon form
RREF = reduced row echelon form

I am currently reading the section on the rank of a matrix and finding basis using row space/column space.

In one book it has me setup a matrix and then solve for REF.
The other book goes all the way and has me to RREF to the same matrix.

Is there a preferred way to do these problems, I mean, obviously doing REF is less work than RREF...

6. Jul 20, 2010

hunt_mat

I've done RREF but I haven't come across REF before (poor education on my part I suspect).
Everyone has their own way of doing problem, there is no preferred way per se, just the persons preferred way.

7. Jul 20, 2010

HallsofIvy

Staff Emeritus
Here's a different, more fundamental way of doing this:

If the five given vectors do not already form a basis for for their span, they must be dependent- there exist numbers a, b, c, d, e, not all 0, such that a<1, 1, 2, 1>+ b<1, 0, -3, 1>+ c<0, 1, 1, 2>+ d<0, 0, 1, 1>+ e<1, 0, 0, 1>= <0, 0, 0, 0>.

That gives us the four equations a+ b+ e= 0, a+ c= 0, 2a- 3b+ c+ d= 0, and a+ b+ 2c+ d+ e= 0. Of course, equation 2 gives c= -a so the third equation is the same as a- 3b+ d= 0 and the last is -a+ b+ d+ e= 0. Since a+ b+ e= 0, b+ e= -a and the last equation becomes -2a+ d= 0 or d= 2a. Putting both c= -a and d= 2a into the 2a- 3b+ c+ d= 0 gives 3a- 3b= 0 or b= a. Finally, then a+ b+ e= 0 becomes 2a+ e= 0 so e= -2a. Our original equation, a<1, 1, 2, 1>+ b<1, 0, -3, 1>+ c<0, 1, 1, 2>+ d<0, 0, 1, 1>+ e<1, 0, 0, 1>= <0, 0, 0, 0>, becomes a<1, 1, 2, 1>+ a<1, 0, -3, 1>- a<0, 1, 1, 2>+ 2a<0, 0, 1, 1>- 2a<1, 0, 0, 1>= <0, 0, 0, 0>. We can divide through by a and then solve for any one of those vectors in terms of the other four- the five given vectors span all of $R^4$ and any four of the original five vectors will be a basis.