# How to prove that the composition $T_2 \circ T_1$ is translation invariant?

Suppose $T_1, T_2 : \ell^2(Z_N)-> \ell^2(Z_N)$ are translation invariant linear transformation. How to prove that the composition $T_2 \circ T_1$ is translation invariant? Recall the translation operator $R_k$ defined by $(R_k z)(n)=z(n-k)$ Also recall definition: Let $T : \ell^2(Z_N)-> \ell^2(Z_N)$ be a linear transformation. $T$ is translation invariant if $T(R_k z)=R_k T(z)$

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$T_2(T_1(R_kz))=T_2(R_kT_1(z))=R_kT_2(T_1(z))$, no? – Raskolnikov Nov 22 '10 at 10:46
I wish you'd at least show what you've tried to solve this... – J. M. Nov 22 '10 at 10:47
Well first $T_2 o T_1(R_k z )$ and then because T_1 is translation invariatnYou get $T_2(R_k(T_1(z ))$. And at last but not least, because T_2 is translation invraiant you get $R_k (T_2((T_1(z ))$. – alvoutila Nov 22 '10 at 11:06
(i) to get the composition operator, use \circ, not o. (ii) You shold tell us what you've tried. You have posted 22 questions, all of them fairly clearly homework assignments. A quick random sampling shows you seldom, if ever, give anything except the statement. Moreover, you have only accepted 4 answers so far even though 18 questions have been answered. This site is not meant to be simply a "help-me-do-my-homework" site, and you are treating people's time rather cavalierly. – Arturo Magidin Nov 22 '10 at 15:46