Let be a linear operator one the finite-dimensional space . Suppose there is a linear operator on such that = . Prove that is invertible and Give an example which shows that this is false when is not finite-dimensional.

THANKS

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- October 2nd 2009, 07:12 PMosodudLinear Transformation Example
Let be a linear operator one the finite-dimensional space . Suppose there is a linear operator on such that = . Prove that is invertible and Give an example which shows that this is false when is not finite-dimensional.

THANKS - October 3rd 2009, 12:33 AMaman_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)

Please check it though

Here V is F[x] - all polynomials in x - October 5th 2009, 07:09 AMtonio

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