# Why are IBM's basis gates not linearly independent?

IBM lists the basis gates for its Heron processor as $$CZ,RZ(\theta),SX,X,I$$. For clarity, these are the Controlled-Z, Z-rotation, sqrt(Pauli-X), Pauli-X, and identity gates respectively.

My understanding is that the native gate set should form a basis. However, this set of gates are not linearly independent since $$RZ(0)=I$$. Can we therefore remove $$I$$ from this set and still have a basis? Alternatively, can we remove $$RZ(\theta)$$ and still have a basis?

However, this does not mean you cannot have gates that can be constructed from each other in your native gate-set. For example, in the gate-set you shared, you can implement an $$X$$ gate using two sequential $$SX$$ gates. So why have a separate $$X$$ gate? Well, because, at the pulse level, there might be a more efficient way to implement a commonly-used gate like the $$X$$ gate than by using two sequential $$SX$$ gates. In the case of IBM's hardware, an $$X$$ gate is implemented to take just as long as an $$SX$$ gate instead of twice as long.
In the case of the $$I$$ gate, you might wonder why even add it since the identity corresponds to not applying any gates at all, making it trivial. However, in some situations you want to keep an $$I$$ as part of the native gate-set in case you need to have a qubit idle for some amount of time. In the case of IBM's hardware that's what the $$I$$ gate does if you don't let the transpiler optimize it out of your circuit.
Lastly, you ask if we could remove $$RZ(\theta)$$ and still have a basis. The answer is no. You need $$RZ$$ to be able to span the majority of the Hilbert space. With only $$CZ$$ and $$SX$$ you can't do that. For example, in the case of a single qubit, how would you implement something as simple as a Hadamard gate with only an $$SX$$ gate? At a minimum, you need a two qubit gate and a pair of single qubit gates dictated by the Solovay-Kitaev theorem (e.g. $$H$$, $$T$$, $$CX$$).