Question about Linear Maps.

PiAreSquared

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.

voko

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.

PiAreSquared

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?

voko

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!

PiAreSquared

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.

voko

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.

PiAreSquared

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.

Bacle2

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.

voko

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.