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Finding the rank of a projection of u onto v ?

  1. Jul 14, 2008 #1
    hello again,

    I'm once again stumped, i was asked to find the rank and nullity of the projection u onto v so here is the given:

    T(u)=ProjvU, where v = <2,4>

    and this is what i did:

    let u = <u1 , u2> and plugged everything in the projection formula and ended up with < 4 + 2(u1) , -16 + 4 (u2)> / sqrt(20)

    how am i supposed to get the rank from that, it is not even a matrix! all i can think of is that a projection diagram looks like a right triangle and i would assosiate that with triangular matricies and triangular matricies have a pivot in every column, therefore the rank should be 2. and therefore nullity is 0 .

    am i on the right track? thanks in advance.
     
  2. jcsd
  3. Jul 16, 2008 #2

    tiny-tim

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    Hi mohdhm! :smile:

    I'm not following you … what are the spaces U and V? :confused:
     
  4. Jul 17, 2008 #3
    i'm sorry mate, u and v are vectors. (and u1 is the x component of u, and u2 is the y component of u, etc.)

    and the question was supposed to be:

    find the rank and nullity of linear transformation T(u) = ProjvU where v is (1,5). i just assumed that u = < u1, u2>.... how do we get a matrix out of that?
     
    Last edited: Jul 17, 2008
  5. Jul 17, 2008 #4

    HallsofIvy

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    That doesn't answer the question- vectors in what vector space? I suspect U and V are both R2 from the way you have written the vectors.

    First, you haven't shown how you got those calculations but the result you give is now where near correct. In particular, NO linear transformation, in particular no projection can have numbers that are not multiplied by u1, u2, the components of the matrix. What I get for the projection of <u1, u2> is <u1+ 2u2, 2u1+ 4u2>/5.

    Second, a projection is a linear transformation and any linear transformation can be written as a matrix. Using the standard basis, <1, 0> and <0, 1>, <1, 0> is projected to <1/5, 2/5> and <0, 1> is projected to <2/5, 4/5>. That corresponds to the matrix
    [tex]\left[\begin{array}{cc}\frac{1}{5} & \frac{2}{5} \\ \frac{2}{5} & \frac{4}{5}\end{array}\right][/tex]
    You idea about how a projection looks like a triangle so its matrix must be triangular makes no sense at all.

    Finally, you don't need to do any of those calculations. the "rank" of a linear transformation, A, on vector space U, is just the dimension of its image, A(U). In the case of a projection onto a single line, that's trivial.
     
  6. Jul 17, 2008 #5
    one last question mate, why do you use the standard basis in this case?
     
  7. Jul 17, 2008 #6

    HallsofIvy

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    Because there was no reason to use any other basis. Another basis would give a matrix equivalent to that (A equivalent to B if and only if there exist an invertible matrix P such that A= PBP-1 or, equivalently, AP= PB. Two matrices are equivalent if and only if they represent the same linear transformation in two different bases.) If I had, for example, used v= <2,4> and <4, -2>, which is orthogonal to v as basis, since the projection of v is just itself and of <4, -2> is the 0 vector, the matrix in that basis is just
    [tex]\left[\begin{array}{cc}1 & 0 \\ 0 & 0\end{array}\right][/tex]

    Do you see that
    [tex]\left[\begin{array}{cc}\frac{1}{5} & \frac{2}{5} \\ \frac{2}{5} & \frac{4}{5}\end{array}\right]\left[\begin{array}{cc}1 & -2 \\ 2 & 1\end{array}\right]= \left[\begin{array}{cc} 1 & -2 \\ 2 & 1\end{array}\right]\left[\begin{array}{cc}1 & 0 \\ 0 & 0\end{array}\right][/tex]

    Can you now answer this: Suppose P(v) is the projection from from R[3] to the plane,through the origin, given by 3x+ 3y+ z= 0. What are the rank and nullity (dimension of kernel) of P?

    You should be able to write them down immediately without doing any calculation.
     
  8. Jul 17, 2008 #7
    yup, solved it. thankyou.
     
  9. Jul 17, 2008 #8

    HallsofIvy

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    So, what was your answer?
     
  10. Jul 17, 2008 #9
    wait answered th wrong question
     
  11. Jul 17, 2008 #10
    the answer is rank = 1 and nullity = 2 for your question
     
  12. Jul 17, 2008 #11

    HallsofIvy

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    No, it's the other way around! Since the projection was onto a plane, the projection maps a 3 dimensional vector into a two dimensional vector. Any vector perpendicular to the plane- that is any vector on the line through (0,0,0) perpendicular to that plane is mapped into the 0 vector and so is in the kernel of the projection. The rank is 2 and the nullity is 1.

    For your original problem, you had a projection from two dimensional space onto a one dimensional subspace, a line. Any vector perpendicular to that, that is any vector parallel to the line perpendicular to the vector <2, 4> is in the null space. Hopefully you got 1 for both rank and nullity.
     
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