Proof of form of 4D rotation matrices I am considering rotations in 4D space. We use $x, y, z, w$ as coordinates in a Cartesian basis. I have found sources that give a parameterization of the rotation matrices as
\begin{align}
    &R_{yz}(\theta) = 
    \begin{pmatrix}
    1&0&0&0\\0&\cos\theta&-\sin\theta&0\\0&\sin\theta&\cos\theta&0\\0&0&0&1
    \end{pmatrix},
    R_{zx}(\theta) = 
    \begin{pmatrix}
    \cos\theta&0&\sin\theta&0\\0&1&0&0\\-\sin\theta&0&\cos\theta&0\\0&0&0&1
    \end{pmatrix},\\
    &R_{xy}(\theta) = 
    \begin{pmatrix}
    \cos\theta&-\sin\theta&0&0\\\sin\theta&\cos\theta&0&0\\0&0&1&0\\0&0&0&1
    \end{pmatrix},
    R_{xw}(\theta) = 
    \begin{pmatrix}
    \cos\theta&0&0&-\sin\theta\\0&1&0&0\\0&0&1&0\\\sin\theta&0&0&\cos\theta
    \end{pmatrix},\\
    &R_{yw}(\theta) = 
    \begin{pmatrix}
    1&0&0&0\\0&\cos\theta&0&-\sin\theta\\0&0&1&0\\0&\sin\theta&0&\cos\theta
    \end{pmatrix},
    R_{zw}(\theta) = 
    \begin{pmatrix}
    1&0&0&0\\0&1&0&0\\0&0&\cos\theta&-\sin\theta\\0&0&\sin\theta&\cos\theta
    \end{pmatrix},
\end{align}
where the subscript labels a plane that is being rotated. This seems to be a very intuitive extension of lower dimensional rotations. However, I would really like to see a proof that these are correct, and I'm not sure how I could go about doing that. By correct, I mean that these 6 matrices can generate any 4D rotation.
My initial attempt was to construct a set of transformations from the definition of the transformations (as matrices) that define a 4D rotation,
\begin{align}
\{R|RR^T = I\},
\end{align}
where $I$ is the identity matrix (4D), but this has 16 (constrained) parameters and I thought that there must be an easier way.
 A: An indirect and sneaky way of doing this is to construct the generators $L_{ij}=-i d R_{ij}/d\theta \vert_{\theta=0}$ and verify that the resulting matrices span the Lie algebra $\mathfrak{so}(4)$. This avoids having to construct a general rotation matrix as a product of your 6 elements.  You can then use the result that
the exponential of any linear combo of your generators is guaranteed to generate an element in the group.
A more direct way is to check that $R_{ij}R_{ij}^T=\mathbb{I}$, and then check that $(R_{ij}R_{ab})^T (R_{ij}R_{ab})=\mathbb{I}$ and then by induction that any product $R=R_{ab}R_{cd}R_{ef}...$ satisfies $RR^T=\mathbb{I}$.
A: Rotation matrices are orthogonal matrices, i.e., $$R^{-1}=R^T.$$
An orthogonal n-by-n matrix has $n(n-1)/2$ independent parameters (see here for an elegant proof), which in our case is 6 parameters. Linear algebra tells us that a 6-component vector can be represented in terms of 6 linearly independent vectors. The rest is to prove that the six given matrices are linearly independent, i.e., that there are no such coefficients $c_j$ that
$$
\sum_jc_jR_j(\theta_j)=0
$$
for arbitrary combination of $\theta_j$, which is easily done by hand.
A: For each of your 4D rotation matrix $~\mathbf R~$ if this equation
$$\mathbf Z^T\, \mathbf Z= \left(\mathbf R\,\mathbf Z\right)^T\,\left(\mathbf R\,\mathbf  Z\right)$$
is fulfilled the rotation matrix $~\mathbf R~$ is orthonormal .$~\mathbf R^T\,\mathbf R=\mathbf I_4$
where
$$\mathbf Z= \begin{bmatrix}
   x \\
   y \\
   z \\
   w \\
 \end{bmatrix}$$
Edit
you can  also  check the determinate of the Rotation matrix ,if the determinate of the Rotation  matrix   is equal one the matrix is orthonormal ?
