1. ## Linear Transformation Example

Let $T$ be a linear operator one the finite-dimensional space $V$. Suppose there is a linear operator $U$on $V$ such that $TU$ = $I$. Prove that $T$ is invertible and $U =T^-1$ Give an example which shows that this is false when $V$ is not finite-dimensional.

THANKS

2. Originally Posted by osodud
Let $T$ be a linear operator one the finite-dimensional space $V$. Suppose there is a linear operator $U$on $V$ such that $TU$ = $I$. Prove that $T$ is invertible and $U =T^-1$ Give an example which shows that this is false when $V$ is not finite-dimensional.

THANKS
TU(v) = T(U(v)) = v

Hence range of T is entire V i.e. T is onto V.
As T is onto AND V is finite dimensional => T is invertible.

TU = I
(T^-1)TU = T^-1
U=T^-1

As for the example in infinite case - can't think. But I guess there was something like
T(q(x)) = q'(x)
U(q(x) = Integral from 1 to x q(x)

Here V is F[x] - all polynomials in x

3. Originally Posted by aman_cc
TU(v) = T(U(v)) = v

Hence range of T is entire V i.e. T is onto V.
As T is onto AND V is finite dimensional => T is invertible.

TU = I
(T^-1)TU = T^-1
U=T^-1

As for the example in infinite case - can't think. But I guess there was something like
T(q(x)) = q'(x)
U(q(x) = Integral from 1 to x q(x)

Here V is F[x] - all polynomials in x

Counterexample in the infinite dimensional case: let V be the real linear space of all infinite sequences and let T be the linear map defined by:

T({a_1, a_2,....,a_n,...}):= {a_2, a_3,....}

If U is defined as U({a_1, a_2,...}):= {0, a_1, a_2,...}, then we get
TU = I , but T isn't invertible since it is very not 1-1 (can you see it?)

Now just check both T, U are certainly linear maps.

Remark: if you prefer to make the example slightly fancier you may even choose V to be the l. s. of all convergent sequences or even all the convergent to zero sequences.

Tonio