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Find all 10x10 Matrices such that ColA=NulA

  1. Dec 14, 2011 #1
    1. The problem statement, all variables and given/known data

    Find all 10x10 Matrices such that the column space is equal to the null space.

    2. Relevant equations

    Choose Function: n!/k!(n-k)!
    where n is the total number of elements and k is the number k-cominations of the set.

    rankA+dimNulA=n for a matrix in R^n

    3. The attempt at a solution
    There is a short and simple proof that we can use to determine that the rank of such a matrix should be 5.

    Proof: Let A be a 10x10 matrix and suppose that ColA=NulA
    => rankA=dimNulA
    => rankA+dimNulA=10
    => rankA+rankA=10
    => 2rankA=10
    => rankA=5

    Ok so from this I can deduce that I will have 5 linearly independent vectors in R^10 that will make up my matrix.

    Now what I was thinking about this is that this must mean that to find ALL matricies I simply have to take permuations of my linearly independent vectors. For example a possibility for a matrix would be <e1,e2,e3,e4,e5,0,0,0,0,0> where e1,...,e5 are the basis vectors for R^10 and those 0's are the 0 vectors in r^10. Another possibility would be
    <0,e2,e4,e5,0,0,0,e1,e3,0> and so on.

    This seems like it'd be A LOT of possibilites so what I thought about doing was utilizing the choose function and from it's utilization I determined that because I have 10 columns total, with 5 vectors as a possibility to make up those columns, then I should have 252 possible combinations which are simply permutations of a 10x10 matrix with the vectors e1,...e5 as a basis.

    Is my line of thinking correct or am I way off?

    Thanks for the help!
     
  2. jcsd
  3. Dec 14, 2011 #2

    Office_Shredder

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    You have more restrictions than just having rank 5. For example, the matrix
    <e1,e2,e3,e4,e5,0,0,0,0,0> has as its null space span(e6,e7,e8,e9,e10) but has as its column space span(e1,e2,e3,e4,e5). If I tell you that the columns have to be e1,e2,e3,e4 and e5 can you identify which columns you can put them in in order to force the condition that the null space and column space are the same?
     
  4. Dec 14, 2011 #3

    micromass

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    You seem to be focussing on finding all the 10x10 matrices that have rank 5. However, not every matrix with rank 5 will satisfy Col(A)=Nul(A). For example

    [tex]\left(\begin{array}{cccccccccc}
    1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\
    0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\
    0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\
    0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0\\
    0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0\\
    0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\
    0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\
    0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\
    0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\
    0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\
    \end{array}\right)[/tex]

    Will not have Col(A)=Nul(A). So you have a lot of redundancy.

    Maybe you should think about what [itex]A^2[/itex] is.
     
  5. Dec 14, 2011 #4
    I'm really confused what do you mean by A^2? Are you saying that A is the identity matrix? I guess I'm just confused in general about WHEN the column is EVEN equal to the null space at all.. How can you tell?
     
  6. Dec 14, 2011 #5

    micromass

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    No, not at all.

    Take an arbitrary x. What can you say about [itex]Ax[/itex] (think about Col(A))? What about [itex]AAx[/itex]. What can you conclude about [itex]A^2[/itex]??
     
  7. Dec 14, 2011 #6
    So what we can say about Ax is that it will be a linear combination of the column vectors of A?
     
  8. Dec 14, 2011 #7

    micromass

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    Yes,so it will be an element of Col(A) rigtht?? But Col(A)=Nul(A), so...
     
  9. Dec 14, 2011 #8
    Ok so the column space of A consists of all possible products Ax for any x an element of R^n. So if the Col(A) is equal to the Nul(A) which is the set of all vectors x for which Ax = 0 then this tells us that Ax must be equal to zero? aka the linear combination must be equal to 0?
     
  10. Dec 14, 2011 #9

    micromass

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    No.

    The only thing you know now is that Ax is in Col(A). So Ax is in Nul(A). So...
     
  11. Dec 14, 2011 #10
    Well if Ax is in Nul(A) then isn't the null space of A the same as the solution set to the homogeneous system Ax=0?
     
  12. Dec 14, 2011 #11
    So for a 10x10 matrix the null space of that matrix would consist of all vectors x in R^10 such that Ax=0 ?
     
  13. Dec 15, 2011 #12

    micromass

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    Yes, y is in the nullspace if Ay=0.

    Now you know that Ax is in the nullspace. So what must hold?
     
  14. Dec 15, 2011 #13
    So Ax must equal 0?
     
  15. Dec 15, 2011 #14

    micromass

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    No. Ax is in the nullspace. That does not mean that it equals 0.
     
  16. Dec 15, 2011 #15
    Ohh ok do you mean that if Ax is in the null space then:
    Nul(A) must contain the zero vector
    If x ∈ Nul(A) and y ∈ Nul(A), then x + y ∈ Nul(A) (aka closed under addition)
    If x ∈ Nul(A) and c is a scalar, then cx ∈ Nul(A) (aka closed under scalar multiplication)
     
  17. Dec 15, 2011 #16

    micromass

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    y is in Nul(A). What does that mean?? What is the definition??

    Now apply this on y=Ax.
     
  18. Dec 15, 2011 #17
    So if a vector y in is Nul(A) then this means that the equation y=c1v1+...+cnvn with cn scalars must equal 0?
     
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