= Solution
\b[The gate set is not universal.] The displayed two-<qubit> <matrix> is the <SWAP gate>:
$$
W(|u\rangle\otimes|v\rangle)=|v\rangle\otimes|u\rangle.
$$
Both $V$ and the <Hadamard gate> act on one <qubit>. Thus every gate preserves a fully factored <product state>, and induction gives the same property for every <quantum circuit> built from them. This is the <product-state obstruction to SWAP universality>.
For a quantitative obstruction to even approximate universality, every two-<qubit> <product state> has squared overlap at most $1/2$ with a <Bell state>. Indeed, for normalized coefficient vectors $u,v$,
$$
|\langle\phi_+|u\otimes v\rangle|^2
=\frac{|u_0v_0+u_1v_1|^2}{2}\leq\frac12
$$
by the <Cauchy-Schwarz inequality>. A <controlled-NOT gate> acting on $|+\rangle|0\rangle$ would instead produce that <Bell state>, so these circuits cannot approximate all <unitary transformations>. The fact that $V^2=e^{-i\pi/8}T$ supplies a <T gate> up to <global phase> does not supply an <entangling gate>.
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