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Question about Linear Maps.

  1. Aug 22, 2012 #1
    Hello,
    I have been working through Linear Algebra Done Right by Axler and I have a quick question about Linear Maps, and in particular, their inverses. My question arose while working through the following proof:

    A linear map is invertible iff it is bijective.

    My qualm is not with the proof itself, as it is quite straightforward. My question however is this: When assuming that a linear map T:V [itex]\rightarrow[/itex] W is bijective and showing that T must then be invertible, the proof first shows that if, for each w [itex]\in[/itex] W, Sw is defined to be the unique element of V that is mapped to w, that is T(Sw) = w, then clearly TS=I and it is shown that ST=I, where the first I is the Identity Map on W and the second is the Identity Map on V. This direction of the proof then finishes by showing that S is linear. But my question is, if S satisfies the properties: ST=I and TS=I, but S is not linear, what is S? Is it just a non-linear map that "undoes" T, because by definition, S cannot be the inverse of T?

    Thanks.
     
  2. jcsd
  3. Aug 22, 2012 #2
    It may be just me, but I don't understand your question. If S satisfies ST = I and T is linear, then S is automatically linear.
     
  4. Aug 22, 2012 #3
    Well, he assumes that T is linear, and shows that ST = I and TS = I, but he then says that to complete the proof, it is necessary to show that S is linear. But if S is automatically linear if T is linear and ST = I, why then does he take the time to show explicitly that S is linear?
     
  5. Aug 22, 2012 #4
    If a map is linear, its inverse is linear, if it exists. The book may show that explicitly, but I still don't understand why you could think S might be non-linear? Especially with an explicit proof of that!
     
  6. Aug 22, 2012 #5
    Yeah that was my point of confusion. I could not figure out how S could be non-linear and still satisfy ST=I and TS=I. So I guess I just assumed that if T was invertible, then its inverse was by default linear, without realizing that it needed to be shown explicitly.

    Sorry for the confusion.
     
  7. Aug 22, 2012 #6
    Some things are intuitively clear, but mathematics must be rigorous (even though intuition is very important). As far a I am concerned, I love proofs of "obvious" properties, they are usually very enlightening.
     
  8. Aug 22, 2012 #7
    voko,

    I have a quick question about one of your previous posts on this thread. You said that "If S satisfies ST = I and T is linear, then S is automatically linear." I was wondering why that is true? I know that If a linear map is invertible, then its inverse is also linear. But for T to be invertible, doesn't S have to satisfy both ST = I AND TS = I? Was this something that was inadvertently left out or does ST = I somehow imply that TS = I and I'm just not seeing it?

    Sorry if I have overlooked another simple fact. I have went through this chapter on linear maps in only a day.
     
  9. Aug 23, 2012 #8

    Bacle2

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    Science Advisor

    Maybe I shouldn't write when it's late and I'm tired, but let's see:

    I don't know, let's see for an argument that the inverseof a linear map is linear:

    Maybe we can start by showing that if the inverse exists, it is unique:

    Assume A is an invertible linear operator on a given choice of basis, with

    AB=AB'=I , then multiply both sides on the left by A-1, to get:

    A-1AB=B , and A-1AB'=B' . Then B=B'.

    Now, given an invertible linear operator A, we can find a linear operator B

    which is its inverse. By above argument , B is _the_ inverse of A, and B is

    linear. So the inverse of a linear operator is linear.
     
  10. Aug 23, 2012 #9
    Note I did not say that if [itex]ST = I[/itex], then [itex]T[/itex] is invertible. I merely said that [itex]S[/itex] is linear if it satisfies that condition and [itex]T[/itex] is linear. Indeed, let [itex]x = Ta[/itex], then [itex]Sx = STa = Ia = a[/itex]; now, [itex]\alpha Sx = \alpha a = I \alpha a = ST \alpha a = S \alpha Ta = S \alpha x[/itex], i.e., [itex]\alpha Sx = S \alpha x[/itex]. Likewise, let [itex]y = Tb[/itex], then [itex]Sy = STb = Ib = b[/itex]; [itex]Sx + Sy = a + b = I(a + b) = ST(a + b) = S(Ta + Tb) = S(x + y)[/itex], i.e., [itex]Sx + Sy = S(x + y)[/itex], so S is linear.
     
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