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Denote by $\DeclareMathOperator{\uso}{\underline{so}}$ $\uso(n)$ the space of symmetric $n\times n$ matrices. The Cayley transform $\newcommand{\eC}{\mathscr{C}}$

$$ \uso(n) \ni X\mapsto \eC(X) := (1-X)(1+X)^{-1} $$

defines an diffeomorphism

$$ \eC:\uso(n)\to SO(n)^*:= \bigl\lbrace T\in SO(N);\;\; \ker(1+T)=0\;\bigr\rbrace. $$

Observe that $SO(n)^*$ is an open an dense subset of $SO(n)$.

Now observe that

$$ \forall X\in\uso(n):\;\;\ker X=\ker\bigl(\; 1-\eC(X)\;\bigr). $$

Thus the space $\uso(n)_k$ consisting of skew symmetric matrices $X\in\uso(n)$ such that $\dim \ker X= k$ has the same dimension as the space of orthogonal $n\times n$ matrices with the property that the eigenspace corresponding to the eigenvalue $1$ has dimension $k$. If $k\equiv n\bmod 2$ we deduce that this space the same dimension as the group $SO(n-k)$.

Hence

$$\dim \uso(n)_k= \binom{n-k}{2}. $$

The space you are interested is $\uso(n)_{n-4}$ and we deduce

$$\dim \uso(n)_{n-4}=\binom{4}{2}= 6. $$

Edit 1. I just realized that invoking the Cayley transform was not really needed. Note that for any $X\in \uso(n) $ we have

$$\dim \ker X equiv n\bmod 2. $$

If $k\equiv n \bmod 2$ we have a diffeo $\uso(n)_k\to\uso(n-k)_0$ which associates to a matrix $X\in\uso(n)_k$ its restriction to the orthogonal complement of the kernel.

Denote by $\DeclareMathOperator{\uso}{\underline{so}}$ $\uso(n)$ the space of symmetric $n\times n$ matrices. $\DeclareMathOperator{\rank}{rank}$ Note that for $A\in \uso(n)$

$$ \rank A=n-\dim\ker A, $$

$$\dim \ker A \equiv n\bmod 2. $$

Denote by $ \uso(n)_k $ the space consisting of matrices $X\in\uso(n)$ such that $\dim \ker X= k$, where $k\equiv n\bmod 2$.

You are interested in the space $\uso(n)_{n-4}$.

We have a diffeo $\uso(n)_k\to\uso(n-k)_0$ which associates to a matrix $X\in\uso(n)_k$ its restriction to the orthogonal complement of the kernel. We deduce

$$\dim \uso(n)_k=\dim \uso(n-k)_0=\dim \uso(n-k)=\binom{n-k}{2}. $$

Thus, in your case

$$\dim \uso(n)_{n-4}=\binom{4}{2}= 6. $$

Denote by $\DeclareMathOperator{\uso}{\underline{so}}$ $\uso(n)$ the space of symmetric $n\times n$ matrices. The Cayley transform $\newcommand{\eC}{\mathscr{C}}$

$$ \uso(n) \ni X\mapsto \eC(X) := (1-X)(1+X)^{-1} $$

defines an diffeomorphism

$$ \eC:\uso(n)\to SO(n)^*:= \bigl\lbrace T\in SO(N);\;\; \ker(1+T)=0\;\bigr\rbrace. $$

Observe that $SO(n)^*$ is an open an dense subset of $SO(n)$.

Now observe that

$$ \forall X\in\uso(n):\;\;\ker X=\ker\bigl(\; 1-\eC(X)\;\bigr). $$

Thus the space $\uso(n)_k$ consisting of skew symmetric matrices $X\in\uso(n)$ such that $\dim \ker X= k$ has the same dimension as the space of orthogonal $n\times n$ matrices with the property that the eigenspace corresponding to the eigenvalue $1$ has dimension $k$. If $k\equiv n\bmod 2$ we deduce that this space the same dimension as the group $SO(n-k)$.

Hence

$$\dim \uso(n)_k= \binom{n-k}{2}. $$

The space you are interested is $\uso(n)_{n-4}$ and we deduce

$$\dim \uso(n)_{n-4}=\binom{4}{2}= 6. $$

Edit 1. I just realized that invoking the Cayley transform was not really needed.

Denote by $\DeclareMathOperator{\uso}{\underline{so}}$ $\uso(n)$ the space of symmetric $n\times n$ matrices. The Cayley transform $\newcommand{\eC}{\mathscr{C}}$

$$ \uso(n) \ni X\mapsto \eC(X) := (1-X)(1+X)^{-1} $$

defines an diffeomorphism

$$ \eC:\uso(n)\to SO(n)^*:= \bigl\lbrace T\in SO(N);\;\; \ker(1+T)=0\;\bigr\rbrace. $$

Observe that $SO(n)^*$ is an open an dense subset of $SO(n)$.

Now observe that

$$ \forall X\in\uso(n):\;\;\ker X=\ker\bigl(\; 1-\eC(X)\;\bigr). $$

Thus the space $\uso(n)_k$ consisting of skew symmetric matrices $X\in\uso(n)$ such that $\dim \ker X= k$ has the same dimension as the space of orthogonal $n\times n$ matrices with the property that the eigenspace corresponding to the eigenvalue $1$ has dimension $k$. If $k\equiv n\bmod 2$ we deduce that this space the same dimension as the group $SO(n-k)$.

Hence

$$\dim \uso(n)_k= \binom{n-k}{2}. $$

The space you are interested is $\uso(n)_{n-4}$ and we deduce

$$\dim \uso(n)_{n-4}=\binom{4}{2}= 6. $$

Edit 1. I just realized that invoking the Cayley transform was not really needed. Note that for any $X\in \uso(n) $ we have

$$\dim \ker X equiv n\bmod 2. $$

If $k\equiv n \bmod 2$ we have a diffeo $\uso(n)_k\to\uso(n-k)_0$ which associates to a matrix $X\in\uso(n)_k$ its restriction to the orthogonal complement of the kernel.

Denote by $\DeclareMathOperator{\uso}{\underline{so}}$ $\uso(n)$ the space of symmetric $n\times n$ matrices. The Cayley transform $\newcommand{\eC}{\mathscr{C}}$

$$ \uso(n) \ni X\mapsto \eC(X) := (1-X)(1+X)^{-1} $$

defines an diffeomorphism

$$ \eC:\uso(n)\to SO(n)^*:= \bigl\lbrace T\in SO(N);\;\; \ker(1+T)=0\;\bigr\rbrace. $$

Observe that $SO(n)^*$ is an open an dense subset of $SO(n)$.

Now observe that

$$ \forall X\in\uso(n):\;\;\ker X=\ker\bigl(\; 1-\eC(X)\;\bigr). $$

Thus the space $\uso(n)_k$ consisting of skew symmetric matrices $X\in\uso(n)$ such that $\dim \ker X= k$ has the same dimension as the space of orthogonal $n\times n$ matrices with the property that the eigen space corresponding to the eigenvalue $1$ has dimension $k$. In turn, this has the same dimension as the group $SO(n-k)$.

Hence

$$\dim \uso(n)_k= \binom{n-k}{2}. $$

The space you are interested is $\uso(n)_{n-4}$ and we deduce

$$\dim \uso(n)_{n-4}=\binom{4}{2}= 6. $$

Denote by $\DeclareMathOperator{\uso}{\underline{so}}$ $\uso(n)$ the space of symmetric $n\times n$ matrices. The Cayley transform $\newcommand{\eC}{\mathscr{C}}$

$$ \uso(n) \ni X\mapsto \eC(X) := (1-X)(1+X)^{-1} $$

defines an diffeomorphism

$$ \eC:\uso(n)\to SO(n)^*:= \bigl\lbrace T\in SO(N);\;\; \ker(1+T)=0\;\bigr\rbrace. $$

Observe that $SO(n)^*$ is an open an dense subset of $SO(n)$.

Now observe that

$$ \forall X\in\uso(n):\;\;\ker X=\ker\bigl(\; 1-\eC(X)\;\bigr). $$

Thus the space $\uso(n)_k$ consisting of skew symmetric matrices $X\in\uso(n)$ such that $\dim \ker X= k$ has the same dimension as the space of orthogonal $n\times n$ matrices with the property that the eigenspace corresponding to the eigenvalue $1$ has dimension $k$. If $k\equiv n\bmod 2$ we deduce that this space the same dimension as the group $SO(n-k)$.

Hence

$$\dim \uso(n)_k= \binom{n-k}{2}. $$

The space you are interested is $\uso(n)_{n-4}$ and we deduce

$$\dim \uso(n)_{n-4}=\binom{4}{2}= 6. $$

Edit 1. I just realized that invoking the Cayley transform was not really needed.