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Linear Transformation rank and nullity

  1. Apr 27, 2007 #1
    1. The problem statement, all variables and given/known data

    Let T: R3 --> R3 be the linear transformation that projects u onto v = (3,0,4)

    Find the rank and nullity of T

    2. Relevant equations

    So let u=(x,y,z)


    3. The attempt at a solution


    So I know that

    T(u) = proj. u onto v

    T(u) = [(3x + 4z)/ 25](3,0,4)

    T(u) = 0

    This is where I get confused, in a simalar example from my text (which skips some steps) leads me to belive that I can set 3x + 4z = 0.

    Is not the nullity(T) = dim(ker(T))?

    So would'nt I have to

    T(u) = [(3x + 4z)/ 25](3,0,4) = (0,0,0)

    Create an augmented matrix and find the nullspace?

    I get a different answer than the book is getting?

    Any ideas?

    Rob
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Apr 27, 2007 #2
    If I understand the question correctly, T is a linear transformation that projects any vector u onto the line that passes through both the origin and v.

    We start with a general vector [tex]\left(\begin{array}{cc}x\\y\\z\end{array}\right)[/tex] and determine its image under the linear transformation.

    Then, we try to detemine the "relationships" between x, y and z that must hold for this vector to belong to the nullspace (ie the image of this vector after the transformation is 0).

    Doing this will give you [tex] \frac{3x+4z}{25} \left(\begin{array}{cc}3\\0\\4\end{array}\right) = \left(\begin{array}{cc}0\\0\\0\end{array}\right) [/tex] (something you have already done). You are right in saying that you will eventually get 3x + 4z = 0.

    Can you get a general form for a vector [tex]\left(\begin{array}{cc}x\\y\\z\end{array}\right)[/tex] that will belong to the nullspace, with the knowledge that 3x + 4z = 0? How many free parameters will we need in the expression?
     
    Last edited: Apr 27, 2007
  4. Apr 28, 2007 #3

    HallsofIvy

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    I would have phrased that a little differently! We are given here that v is a vector (because it was boldface!), not a point. I would say that T is the linear transformation that projects any vector u onto the subspace generated by v. (Yes, there is plenty of room for argument- I only said I would "phrase" it differently!)

    There is no need to work out the matrix form for T to answer this question: T maps all of R3 onto a one-dimensional subspace: its rank is 1.
    Of course, then, its nullity is 3- 1= 2.


    Simlarly, if the problem had said that T maps any vector u onto the subspace spanned by v= (3, 0, 4) and w= (1, 1, 1), since those two vectors are independent and span a 2 dimensional subspace, we would have rank of T= 2 and nullity= 3-2= 1.
     
  5. Apr 30, 2007 #4
    So then

    basis for Ker(T) = (x, y, z) = (-4/3, 0, 1) Right?

    Nullity(T) = dim(kernel) = 1

    Rank(T) + Ker(T) = N
    Rank(T) = 3 - 1 = 2

    These are the formulas from my book, do I have the right basis for my kernel then?
     
    Last edited: Apr 30, 2007
  6. May 1, 2007 #5
    There is no restriction on the value of y, because the only condition we have about x, y and z is 3x + 4z = 0. Hence, a free parameter should be assigned to y. Can you figure out what the second basis vector for the nullspace is?

    Since the question did not explicitly ask for the basis vectors of the nullspace, you may want to consider HallsofIvy's approach to the question. And yes, I guess I should have been more careful with the phrasing of my previous post.:approve: Perhaps it should be "T is a linear transformation that projects any vector u onto the line that passes through both the origin and the point with position vector v"?
     
    Last edited: May 1, 2007
  7. May 1, 2007 #6

    HallsofIvy

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    No, that is not the kernel.
    If T projects any vector onto the subspace spanned by (3, 0, 4), the kernel is the set of vectors in R3 perpendicular to (3, 0, 4): the set of vectors (x,y,z) such that (x,y,z).(3,0,4)= 3x+ 4z= 0. Then we must have z= -(3/4)x and all such vectors are of the form (x, y, -(3/4)x). If we take x= 4, y=0, that gives (4, 0, -3) as a vector in that space. If we take x=0, y= 1, that gives (0, 1, 0). Those two vectors are independent (we guarenteed that by taking y= 0 in the first and x= 0 in the second) and form a basis for the kernel.

    T projects every vector v onto the subspace spanned by (3, 0, 4). That subspace, since it is spanned by a single vector, has dimension 1 and so the rank is 1. From that you can calculate that the kernel has dimension 3-1= 2.
     
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