Show that T^kU = UT^k

Nona · #1 $Old$ November 6th, 2009, 08:45 AM

Hello,
i would like help with this:
Assume that T : V ....> V is linear and that n and k are positive integers and that a_0,a_1,a_2,.....,a_n are scalars. Let U = a_0IV + a_1T + a_2T^2 + + a_nT^n. Show that T^kU = UT^k.
Use from 1 to 6:

(1) Since

L(V, V ) is a vector space all of the vector space properties hold for addition and scalar multiplication.
(2) T(U1U2) = (TU1)U2.
(3) ) T(U1 + U2) = (TU1) + (TU2).
(4) (U1 + U2)T = U1T + U2T.
(5) For any non-negative integers i and j, T^iT^j = T^i+j .

(6) For any scalars c and d, (cT )(dU1) = (cd)TU1.

Thank you

tonio · #2 $Old$ November 7th, 2009, 05:31 AM

Quote:

Originally Posted by Nona $View Post$

Hello,
i would like help with this:
Assume that T : V ....> V is linear and that n and k are positive integers and that a_0,a_1,a_2,.....,a_n are scalars. Let U = a_0IV + a_1T + a_2T^2 + + a_nT^n. Show that T^kU = UT^k.
Use from 1 to 6:

(1) Since

L(V, V ) is a vector space all of the vector space properties hold for addition and scalar multiplication.

(2)

T(U1U2) = (TU1)U2.

(3) )

T(U1 + U2) = (TU1) + (TU2).

(4) (U1 + U2)T = U1T + U2T.

(5) For any non-negative integers

i and j, T^iT^j = T^i+j .
(6) For any scalars c and d, (cT )(dU1) = (cd)TU1.

Thank you

This is trivial using the given definitions. For example, with k = 2 we get:

$T^2U=T^2(a_0I+a_1T+...+a_nT^n)=T\left[T\left(\sum\limits_{i=0}^na_iT^i\right)\right]=$ $T\left(\sum\limits_{i=0}^na_iT^{i+1}\right)=\sum\limits_{i=0}^na_iT^{i+2}$

$UT^2=(UT)T=\left[\left(\sum\limits_{i=0}^na_iT^{i}\right)T\right]T$ $=\left[\sum\limits_{i=0}^na_iT^{i+1}\right]T=\sum\limits_{i=0}^na_iT^{i+2}$

Now just apply a simple induction here and you're done.

Tonio

Nona · #3 $Old$ November 7th, 2009, 09:58 AM

Thank you very much.
Is that means: changing 2 to k.
I have to use all the properties from 1 to 6

tonio · #4 $Old$ November 7th, 2009, 10:16 AM

Quote:

Originally Posted by Nona $View Post$

Thank you very much.
Is that means: changing 2 to k.
I have to use all the properties from 1 to 6

Well, in fact you need them more than anything else to fully justify the first case which we already did above. For the case k you must assume that it is true for k-1 and then use induction...and yes, still some of the properties 1-6 given.

Tonio

Nona · #5 $Old$ November 7th, 2009, 02:20 PM

Thank you very much.
Will try to do it.