su(2) Lie algebra elements I have the following three 2 $\times$ 2 complex matrices
\begin{equation}
I_{xx}=\frac{i}{2}\begin{pmatrix} 0 & i \cr -i & 0 \end{pmatrix},\\
I_{yy}=\frac{-i}{2\sqrt{j^2-g^2}}\begin{pmatrix} -j & ig \cr ig & j \end{pmatrix},\\
I_{zz}=\frac{i}{2\sqrt{j^2-g^2}}\begin{pmatrix} ig & j \cr j & -ig \end{pmatrix},
\end{equation}
where $j>g$ are real numbers, $i=\sqrt{-1}$.
It is easy to check that they satisfy the commutation relation (and the Jacobi identity)
\begin{equation}
\begin{aligned}
I_{xx}I_{yy}-I_{yy}I_{xx}=I_{zz}, \\
I_{yy}I_{zz}-I_{zz}I_{yy}=I_{xx}, \\
I_{zz}I_{xx}-I_{xx}I_{zz}=I_{yy}.
\end{aligned}
\end{equation}
I have the following two contradictory belifs:
Belief (1): This implies $I_{xx}$, $I_{yy}$, $I_{zz}$ form a basis of the su(2) Lie algebra. As the commutation relation is the same as
\begin{equation}
\begin{aligned}
u_1=\frac{1}{2}\begin{pmatrix} 0 & i \cr i & 0 \end{pmatrix},\\
u_2=\frac{1}{2}\begin{pmatrix} 0 & -1 \cr 1 & 0 \end{pmatrix},\\
u_3=\frac{1}{2}\begin{pmatrix} i & 0 \cr 0 & -i \end{pmatrix},
\end{aligned}
\end{equation}
which is known to form a basis of su(2) Lie algebra (https://en.wikipedia.org/wiki/Representation_theory_of_SU(2)).
Belief (2): $I_{xx}$, $I_{yy}$, $I_{zz}$ don't form a basis of su(2) Lie algebra, as $I_{yy}$ cannot be written as a linear combination of $u_{1,2,3}$.
Question 1: Which belief is correct (i.e., do $I_{xx}$, $I_{yy}$, $I_{zz}$ form a basis of su(2))? What is wrong about the other?
Question 2: Can $I_{xx}$, $I_{yy}$, $I_{zz}$ after exponentiation form an SU(2) Lie group?
 A: What you have found is a family of subalgebras which are isomorphic to $\mathfrak{su}(2)$. Traditionally, we consider the $\mathfrak{su}(2)$ to be the trace-free linear maps which are skew-self-adjoint for a certain complex inner product (or in matrix terms, skew-Hermitian with trace zero) but by changing the inner product we get different subgroups/subalgebras which are all isomorphic.
So to answer the question: $I_{xx},I_{yy},I_{zz}$ form the basis of a copy of $\mathfrak{su}(2)$ (and changing $j,g$ will give different copies) but they don't have to be the usual copy.
A: You made a mistake. The complex hyperbolic (1-3) rotation coefficients involved convert your antihermitean generators u to the I s which are no longer antihermitian; but there are lost of such bases used for su(2)...
Define $c\equiv 1/\sqrt{j^2-g^2}$ to then readily observe
$$
I_{xx}= u_2,\\
I_{yy}=cj u_3-icg u_1, \\
I_{zz}=icgu_3+cju_1,
$$
and confirm the Lie algebra holds in the new basis, assuming it does in the standard basis.
So your belief (1) is sound, while belief (2) is unsound.
For question 2), the answer is "yes", even though your rotation generators will be ugly non-unitary matrices.
It is a standard homework problem to find the similarity transformation achieving this basis change.
