how to find linearly independent vectors

[love_electronic]

I want to find linearly independent columns of the following matrix and also the nullities of matrix


the property of linearly independent is as follows


can anyone tell me how to find that?

 

[albbg]

I'll try to answer your question even if I'm out of practice for thirty years with matrix math. So, please check carefully the correctness..

If you want to know how many columns are linearly independent you can calculate the rank of he matrix simply looking for the minors (square submatrices) that have not null determinat and counting its rows (or columns). In you example we have (I'll use dots to spearate the minors):

0 1 1.....0 1 2.....1 1 2
1 2 3.....1 3 4.....2 3 4
2 0 2.....2 2 0.....0 2 0

all of them have det=0 the rank<3
doing the same check for the minor 2x2 we will find many having det<>0, then rank=2
Since the rank-nulling theorem says that Rank(A)+Null(A)=n (number of columns) then in our case Null(A)=4-2=2

Or, to find the linearly independent columns you can apply the definition, then calling x1, x2, x3 and x4 the constants that multiply each column we'll have:

0*x1+x2+x3+2x4=0
x1+2x2+3x3+4x4=0
2x1+0x2+2x3+0x4=0

from the last x1=-x3, then substituting:

x2+x3+2x4=0
2(x2+x3)+4x4=0

but from the first x2+x3=-2x4, then the last is alway zero regardless the value of x4 so it's free
now x1 depends from x3 (we saw x1=-x3) but also x2 depends from x3 (and from x4), then we can conclude there are two free variables (x3 and x4) and two dependent variables (x1 and x2) so, again, the rank is 2.

 

[sriblue7]

To find the linearly independent columns perform row echelon operation and look for the pivots.The pivot columns corresponds to linearly independent columns.The number of non-pivot columns indicates the dimension of null space.For deeper understanding i would suggest books by Gilbert strang and his MIT OCW lectures.
For the Matrix you have given the first two columns are linearly independent.The null space of the matrix is all possible linear combinations of [-1 -1 1 0]' and [0 -2 0 0 1]'.I used the commands rref(A) and null(A,'r') in matlab to find the above solutions. A -> given matrix.
Hope this helps.....

 

[love_electronic]

Thanks guys
but i have some queries

all of them have det=0 the rank<3

what about rank if one of det is not equal to 0?

minor 2x2 we will find many having det<>0, then rank=2

which one are 2x2 and how to find its rank?

maybe these are beginners questions but i don't know guys.

 

[albbg]

OK, if at least one of the JxJ minors have determinant <> 0 then the rank is J. in our case all 3x3 minors' determinant are 0 then the rank must be < 3.
some of the 2x2 minors are (you can easily find all the others):

0 1....0 1....0 2....0 1....0 1....0 2....1 2.............
1 2....1 3....1 4....2 0....2 2....2 0....2 4.............

If just one has determinant <> 0 then the rank is 2. As you can see all the 2x2 minors I've written, except the last one, have determinant <> 0 then the rank is 2.

 

[love_electronic]

Thanks albbg. I got the idea, but its a lengthy process.should i use this method or the equation method?

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one more question albbg if i have a matrix
A= 0 1 1 2 -4
1 2 3 4 -8
2 0 2 0 0

then how to find the rank?

 

[albbg]

Yes it's a quite boring process; sorry but I don't remember if passing through the echelon row matrix (as suggested by sriblue7) can be faster.


About the additional question, you add the last column then we have the following "new" 3x3 minors:

0 1 -4....0 1 -4....0 2 -4....1 1 -4....1 2 -4....1 2 -4
1 2 -8....1 3 -8....1 4 -8....2 3 -8....2 4 -8....3 4 -8
2 0 0....2 2 0....2 0 0....0 2 0....0 0 0....2 0 0

All of them have determinant=0 then the rank is always 2.