Introduce a standard scalar product in $\mathbb{R}^4.$ Let $V$ be a partial vector space in $\mathbb{R}^4$ produced by $\left(
\begin{array}{c}
1  \\
-1  \\
-1  \\
1 
\end{array}
\right),\left(
 \begin{array}{c}
1  \\-1  \\
1  \\
-1 
\end{array}
\right).$

Find a pair of base of orthogonal complement $W$ for $V$ in $\mathbb{R}^4.$
1. **Define the subspace \( V \) generated by the given vectors:**

   The given vectors are:
   \[
   \mathbf{v}_1 = \begin{pmatrix} 1 \\ -1 \\ -1 \\ 1 \end{pmatrix}, \quad \mathbf{v}_2 = \begin{pmatrix} 1 \\ -1 \\ 1 \\ -1 \end{pmatrix}
   \]
   The subspace \( V \) generated by these vectors is:
   \[
   V = \text{span}\left\{ \mathbf{v}_1, \mathbf{v}_2 \right\}
   \]

2. **Express the general form of vectors in \( V \):**

   Any vector in \( V \) can be written as a linear combination of \( \mathbf{v}_1 \) and \( \mathbf{v}_2 \):
   \[
   \mathbf{v} = a \mathbf{v}_1 + b \mathbf{v}_2 = a \begin{pmatrix} 1 \\ -1 \\ -1 \\ 1 \end{pmatrix} + b \begin{pmatrix} 1 \\ -1 \\ 1 \\ -1 \end{pmatrix}
   \]
   Simplifying this, we get:
   \[
   \mathbf{v} = \begin{pmatrix} a + b \\ -a - b \\ -a + b \\ a - b \end{pmatrix}
   \]

3. **Find the orthogonal complement \( W \) of \( V \):**

   The orthogonal complement \( W \) consists of all vectors in \( \mathbb{R}^4 \) that are orthogonal to every vector in \( V \). Let \( \mathbf{w} = \begin{pmatrix} w_1 \\ w_2 \\ w_3 \\ w_4 \end{pmatrix} \) be a vector in \( W \). Then \( \mathbf{w} \) must satisfy:
   \[
   \mathbf{w} \cdot \mathbf{v}_1 = 0 \quad \text{and} \quad \mathbf{w} \cdot \mathbf{v}_2 = 0
   \]
   This gives us the following system of equations:
   \[
   \begin{pmatrix} w_1 \\ w_2 \\ w_3 \\ w_4 \end{pmatrix} \cdot \begin{pmatrix} 1 \\ -1 \\ -1 \\ 1 \end{pmatrix} = 0 \quad \Rightarrow \quad w_1 - w_2 - w_3 + w_4 = 0
   \]
   \[
   \begin{pmatrix} w_1 \\ w_2 \\ w_3 \\ w_4 \end{pmatrix} \cdot \begin{pmatrix} 1 \\ -1 \\ 1 \\ -1 \end{pmatrix} = 0 \quad \Rightarrow \quad w_1 - w_2 + w_3 - w_4 = 0
   \]

4. **Solve the system of equations:**

   We have the system:
   \[
   \begin{cases}
   w_1 - w_2 - w_3 + w_4 = 0 \\
   w_1 - w_2 + w_3 - w_4 = 0
   \end{cases}
   \]
   Adding these two equations, we get:
   \[
   2w_1 - 2w_2 = 0 \quad \Rightarrow \quad w_1 = w_2
   \]
   Subtracting the second equation from the first, we get:
   \[
   -2w_3 + 2w_4 = 0 \quad \Rightarrow \quad w_3 = w_4
   \]
   Therefore, any vector \( \mathbf{w} \) in \( W \) can be written as:
   \[
   \mathbf{w} = \begin{pmatrix} w_1 \\ w_1 \\ w_3 \\ w_3 \end{pmatrix} = w_1 \begin{pmatrix} 1 \\ 1 \\ 0 \\ 0 \end{pmatrix} + w_3 \begin{pmatrix} 0 \\ 0 \\ 1 \\ 1 \end{pmatrix}
   \]

5. **Identify the basis for \( W \):**

   The vectors \( \begin{pmatrix} 1 \\ 1 \\ 0 \\ 0 \end{pmatrix} \) and \( \begin{pmatrix} 0 \\ 0 \\ 1 \\ 1 \end{pmatrix} \) are linearly independent and span \( W \). Therefore, they form a basis for \( W \).

The final answer is \( \boxed{ \left\{ \begin{pmatrix} 1 \\ 1 \\ 0 \\ 0 \end{pmatrix}, \begin{pmatrix} 0 \\ 0 \\ 1 \\ 1 \end{pmatrix} \right\} } \).