Invertible linear transformation

[yifli]

Homework Statement

State why it is that if V is finite-dimensional, and if S and T in HomV satisfy S\circ T = I, then T is invertible and S=T^-1



2. The attempt at a solution
Isn't this obvious? I don't understand why any elaboration is necessary

 

[micromass]

Hi yifli!

No, this isn't obvious. T invertible means

S\circ T=1~\text{and}~T\circ S=1.

Now you only have that S\circ T=1. Somehow, you need to show T\circ S=1.

(Hint: use determinants)

 

[jbunniii]

You don't need to use determinants (or even matrices), and in fact doing so might introduce circular logic depending on what theorems you use.

Instead, answer this:

S\circ T = I

implies that T is _____ and S is _____

(Fill in the blanks with either injective or surjective as appropriate.)

Then you'll need to use the finite dimensionality of V to draw a conclusion.

 

[yifli]

You don't need to use determinants (or even matrices), and in fact doing so might introduce circular logic depending on what theorems you use.

Instead, answer this:

S\circ T = I

implies that T is _____ and S is _____

(Fill in the blanks with either injective or surjective as appropriate.)

Then you'll need to use the finite dimensionality of V to draw a conclusion.

implies T is injective and S is surjective
since V is finite dimensional, T is an isomorphism,so S = T^-1

 

[yifli]

Hi yifli!

No, this isn't obvious. T invertible means

S\circ T=1~\text{and}~T\circ S=1.

Now you only have that S\circ T=1. Somehow, you need to show T\circ S=1.

(Hint: use determinants)

Can you give me an example that shows T is not invertible if only ST=I is given?

thanks

 

[jbunniii]

Can you give me an example that shows T is not invertible if only ST=I is given?

thanks

It has to be an infinite-dimensional example, as you just showed.

Let V be the vector space consisting of infinite sequences of the form

(a_1, a_2, a_3, \ldots)

where the a_i are real numbers. (Or complex. It doesn't matter.)

Addition and scalar multiplication are defined pointwise in the obvious way.

Let T be the operator that shifts the sequence right by one step:

T(a_1, a_2, a_3, \ldots) = (0, a_1, a_2, \ldots)

And let S be the operator that shifts the sequence left by one step:

S(a_1, a_2, a_3, \ldots) = (a_2, a_3, a_4, \ldots)

You can verify that these are linear operators, and that

S \circ T = I

but

T \circ S \neq I

T can be "undone" but S cannot.

 

[micromass]

Can you give me an example that shows T is not invertible if only ST=I is given?

thanks

With other things than vector spaces you mean?
Well, take T:{1}-> {1,2} with T(1)=1 and take S:{1,2}->{1} with S(1)=1 and S(2)=1.