Given that $A$ is skew-symmetric matrix of order $n.$,then $A^{T}$ is also be order $n.$

We know that for skew-symmetric $A^{T}=-A$

$\Rightarrow A^{T}+A=[ 0 ]$

Let's take order $n=3$

$A=\begin{bmatrix} 0 &1 &-2 \\ -1 &0 &3 \\ 2 &-3 &0 \end{bmatrix}$

$A^{T}=\begin{bmatrix} 0 &-1 &2 \\ 1 &0 &-3 \\ -2 &3 &0 \end{bmatrix}$

$A+A^{T}=\begin{bmatrix} 0 &1 &-2 \\ -1 &0 &3 \\ 2 &-3 &0 \end{bmatrix}+\begin{bmatrix} 0 &-1 &2 \\ 1 &0 &-3 \\ -2 &3 &0 \end{bmatrix}$

$A+A^{T}=\begin{bmatrix} 0 &0 &0 \\ 0 &0 &0 \\ 0 &0 &0 \end{bmatrix}$

Charactricstic equation for matrix $A+A^{T}:$

$|(A+A^{T})-\lambda I|=0$ where $I=\begin{bmatrix} 1 &0 &0 \\ 0 &1 &0 \\ 0 &0 &1 \end{bmatrix}$ Identity matrix.

$\Rightarrow \begin{vmatrix} -\lambda & 0 &0 \\ 0 & -\lambda &0 \\ 0&0 & -\lambda \end{vmatrix}=0$ where $\lambda$ is the Eigen Value.

$\Rightarrow -\lambda(\lambda^{2}-0)=0$

$\Rightarrow -\lambda^{3}=0$

$\Rightarrow \lambda^{3}=0$

So,Eigen values are $\lambda_{1}=0,\lambda_{2}=0,\lambda_{3}=0$

We know that $AX=\lambda X$

$\Rightarrow(AX-\lambda X)=[0]$

$\Rightarrow((A+A^{T})X-\lambda X)=[0]$ In this question $A=A+A^{T}$

$\Rightarrow((A+A^{T})-\lambda I)X=[0]$ ------->(1)

Now,$(A+A^{T})-\lambda I=\begin{bmatrix} 0 &0 &0 \\ 0&0 &0 \\ 0 & 0 &0 \end{bmatrix}-0.\begin{bmatrix} 1 &0 &0 \\ 0&1 &0 \\ 0 & 0 &1 \end{bmatrix}$ where $\lambda=0$ and $I$ is the identity matrix.

$(A+A^{T})-\lambda I=\begin{bmatrix} 0 &0 &0 \\ 0&0 &0 \\ 0 & 0 &0 \end{bmatrix}-\begin{bmatrix} 0 &0 &0 \\ 0&0 &0 \\ 0 & 0 &0 \end{bmatrix}$ where $\lambda=0$ and $I$ is the identity matrix.

$(A+A^{T})-\lambda I=\begin{bmatrix} 0 &0 &0 \\ 0&0 &0 \\ 0 & 0 &0 \end{bmatrix}$

Now from the equation $(1),$

$((A+A^{T})-\lambda I)X=[ 0 ]$

$\Rightarrow\begin{bmatrix} 0 &0 &0 \\ 0&0 &0 \\ 0 & 0 &0 \end{bmatrix}.\begin{bmatrix} x\\y \\ z \end{bmatrix}=\begin{bmatrix} 0\\0 \\ 0 \end{bmatrix}$ ,where $X=\begin{bmatrix} 0\\0 \\ 0 \end{bmatrix}$ Eigen Vector

Rank$=$Order( Dimension of the matrix)$-$Nullity

Rank of a matrix $A+A^{T} =0$ and number of unknowns(variables) are $3$.

So$,r=0,n=3$

clearly see $r<n$ This is the condition for Infinite many numbers of solutions.

we can assign $n-r=3-0=3$ linealrly independent values to the $3$ unknowns(variables).

let say $x=k_{1},y=k_{2},z=k_{3}$

So,The number of Linealrly independent Eigen Vector is $3$.