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As I know, we can decompose $U$ without ancilla if it's from special unitary group $SU(2^n)$. Do we need to use ancilla qubit on decomposing arbitrary $n$-qubit $U$ using Clifford+T universal gates set?

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No, you don't need ancillae. Ancilla qubits can be extremely helpful, but they are not necessary. For example, in Nielsen and Chuang this is proven by giving an explicit construction.

Also, there's no restriction that the unitary has to be from the special unitary group. Any unitary operation can be approximated to arbitrarily good accuracy using Clifford+T.

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