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Linear Algebra: linear transformation

  1. Mar 13, 2014 #1
    1. The problem statement, all variables and given/known data
    We have seen that the linear transformation ##T(x_1,x_2)=(x_1,0)## on ##\mathcal{R}^2## has the matrix ##A = \left( \begin{smallmatrix} 1&0\\ 0&0 \end{smallmatrix} \right)## with respect to the standard basis. This operator satisfies ##T^2=T##. Prove that if ##S:\mathcal{R}^2 \to \mathcal{R}^2##, is a linear transformation satisfying ##S^2=S##, then ##S=0## or ##S=I## or there exists a basis ##\mathcal{B}## for ##R^2## such that the matrix of ##S## with respect to ##\mathcal{B}## is ##A##.



    2. Relevant equations
    The professor would prefer if we used basic principals to solve this since it is a undergraduate course.

    This homework is past due and I will not get credit for this but it might be on the test so I would appreciate the typical, "make me think" type of replies.


    3. The attempt at a solution
    The fact that ##S^2=S## when ##S=0## or ##S=I## is obvious. However, looking at the dimension of the kernel and the range of ##S## in those cases gives a hint on how to solve the remaining case.
    1st Case: ##S=0\implies## the ##dim(n(S))=2## and ##dim(range(S))=0##
    2nd Case: ##S=I\implies## the ##dim(n(S))=0## and ##dim(range(S))=2## (I has an inverse and therefore is a bijection so the kernel is {0})

    The final case must have that ##dim(n(S))=1=dim(range(S))## since that is the only option left to us.
    ...This is as far as I got.
     
  2. jcsd
  3. Mar 13, 2014 #2

    Dick

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    Yes, ##S^2=S## is obviously satisfied by 0 and I. But you've still got a little work to do. Can you show if dim(range(S))=0 then S must be 0. And a little harder, can you show if dim(range(S))=2 then S must be I? You'll have to use the properties of S to show that. And in the case dim(range(S))=1 there is a single vector v which spans range(S). So? Try to use that to pick a basis.
     
  4. Mar 13, 2014 #3
    ##dim(range(S))=0 \implies## every element of ##R^2## is in the kernel. Using linearity, ##S(v_1+v_2)=S(v_1)+S(v_2)=0=S(R^2)## for all ##v\in R^2##(seems like there should be a better way to do this but it works)

    Suppose ##M(S)=\left (\begin{matrix}a&b\\c&d\end{matrix} \right)## then ##S^2=S \implies ##
    ##a^2+bc=a##
    ##ab+bd=b##
    ##ac+dc=c##
    ##d^2+bc=d##
    Since the dim(range(S))=2 the null space is empty and there is no trivial solution to the above system. This forces a=d=1 and b=c=0 ##\implies S=I##.

    For the final case ##dim(n(S))=1=dim(range(S))##.

    Since the kernel has dimension 1, vectors that get sent to 0 lie on a line in ##R^2##. Vectors that get sent to the range also lie on a line and are of the form ##v_1=cv_2## for any ##v_1,v_2 \in R^2##....not real sure where I'm going with this...need a push in the right direction.

    Edit: I think I made some progress.
    For vectors on the range
    ##S^2=S\implies SSv_1=Sv_1\implies Sv_1=v_1##
    For vectors in the kernel
    ##SSv_2=Sv_2=0##
    And so we want ##v_1## and ##v_2## with
    ##Sv_1=v_1## and ##Sv_2=0##
     
    Last edited: Mar 14, 2014
  5. Mar 14, 2014 #4

    Dick

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    That's getting there. But it's not generally (for any S) true that ##SSv=Sv \implies Sv=v##. What is true for this S, is that if v is in the range then ##v=Su## for some vector u. That means ##Sv=SSu=Su=v##. And you know every v can be written as the sum of a vector in the range plus a vector in the kernel. Use that to think about a nice way to pick a basis.
     
  6. Mar 14, 2014 #5
    ##Sv=SSu=Su=v\implies Sv=kv## for all ##v\in R^2## that get mapped to the range. Where I used the fact that the dimension for the range are all points on a line.
    ##\mathcal{B}={<k,0>,<0,0>}##
    ##S=kA## ?
    If correct, do I need to show this the other way too? Do I need to show that if the basis for the range is 1 dimensional then so is the basis for the vectors that get mapped to it?

    Edit: added little more and I changed the variable for the constant to k to avoid confusion.

    Let ##w_1## be in the set of vectors that get mapped to the range and ##w_2\in kernel(S)##.
    Then##S(w_1+w_2)=S(w_1)+S(w_2)=cw_1+0=cw_1## What does this show?
     
    Last edited: Mar 14, 2014
  7. Mar 14, 2014 #6

    Dick

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    Forget about the line thing. The dimension of a space is the number of vectors in a basis. That's all. Don't "Let ##w_1## be in the set of vectors that get mapped to the range". That's ALL vectors. Let ##w_1## be a nonzero vector in the range. So ##Sw_1=w_1##. Let ##w_2## be a nonzero vector in the kernel. Can you see why ##\{w_1, w_2\}## is a basis for the whole space?
     
  8. Mar 14, 2014 #7

    pasmith

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    We are given that [itex]S: \mathbb{R}^2 \to \mathbb{R}^2[/itex] is linear with [itex]S^2(v) = S(v)[/itex] for all [itex]v \in \mathbb{R}^2[/itex], and we are assuming that [itex]S[/itex] is neither the zero map nor the identity map.

    Since by assumption [itex]S[/itex] is not the zero map, there exists [itex]v \neq 0[/itex] such that [itex]u = S(v) \neq 0[/itex]. But we have [itex]S^2 (v) = S(v)[/itex], ie. [itex]S(u) = u \neq 0[/itex].

    Since by assumption [itex]S[/itex] is not the identity map, there exists [itex]x \neq 0[/itex] such that [itex]S(x) - x \neq 0[/itex]. Yet [itex]S(S(x) - x) = S^2(x) - S(x) = 0[/itex], so there exists [itex]w \neq 0[/itex] such that [itex]S(w) = 0[/itex].

    I claim that [itex]\{u, w\}[/itex], where [itex]S(u) = u[/itex] and [itex]S(w) = 0[/itex], is a basis of [itex]\mathbb{R}^2[/itex].
     
  9. Mar 14, 2014 #8
    "Can you see why ##\{w_1, w_2\}## is a basis for the whole space?"
    Because ##dimension(R^2)=dimension(kernel(S))+dimension(range(S))=1+1=2##
    Edit: Oh wait, they must be linearly independent. I'll check.
    Edit 2: To show L.I., for any two constants ##a,b## we must have:
    ##aS(w_1)+bS(w_2)=0##
    ##aw_1+b*0=0\implies a=0##
    but
    ##aw_1+bw_2=0*w_1+b*w_2=0\implies b=0##
    and so ##\{w_1, w_2\}## are two L.I. vectors in ##R^2## forming a basis and satisfying S^2=S.
     
    Last edited: Mar 14, 2014
  10. Mar 14, 2014 #9

    Dick

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    Good! Now what's the matrix of ##S## is the basis ##\{w_1, w_2\}##?
     
  11. Mar 14, 2014 #10
    Just to finish up. ##A## falls in this category because ##A(1,0)^t=(1,0)^t## and ##A(0,1)^t=0##
    Edit: not quite finished actually.
    Edit2: OK now I'm done!

    ##Sv_1=v_1=a_1v_1+b_1v_2 ##
    ##Sv_2=0=a_2v_1+b_2v_2 ##
    and so S=A.
     
    Last edited: Mar 14, 2014
  12. Mar 14, 2014 #11

    Dick

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    Now you're done. You might want to take that same approach to show S=0 or S=I are the only other possibilities.
     
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