Q:

Understanding direct sum of matrices

I read the definition of direct sum on wikipedia, and got the idea that a direct sum of two matrices is a block diagonal matrix. 
However this does not help me understand this statement in a book. In the book I am reading, the matrix $$
\begin{pmatrix}
0&0&0&1 \\
0&0&1&0 \\
0&1&0&0 \\
1&0&0&0
\end{pmatrix}
$$
"can be regarded as the direct sum of two submatrices":
$$
\begin{pmatrix}
0&1 \\
1&0 
\end{pmatrix},\begin{pmatrix}
0&1 \\
1&0 
\end{pmatrix}$$
Where onen lies in the first and fourth rows (columns) and the other in the second and third. 
According to the definition it should be
$$
\begin{pmatrix}
0&1&0&0 \\
1&0&0&0 \\
0&0&0&1 \\
0&0&1&0
\end{pmatrix}
$$

This was taken from a problem in Problems and Solutions in Group Theory for Physicists by Zhong-Qi Ma and Xiao-Yan Gu. Here's the problem and the solution in full. 

Problem 3. Calculate the eigenvalues and eigenvectors of the matrix $R$
  $$
R = 
\begin{pmatrix}
0&0&0&1 \\
0&0&1&0 \\
0&1&0&0 \\
1&0&0&0
\end{pmatrix}.
$$
Solution. $R$ can be regarded as the direct sum of the two submatrices $\sigma_1$, one lies in the first and fourth rows(columns), the other in the second and third rows(columns). From the result of Problem 2, two eigenvalues of $R$ are $1$, the remaining two are $-1$. The relative eigenvalues are as follows:
  $$
1:  
\begin{pmatrix}
1\\ 0 \\ 0 \\ 1 
\end{pmatrix},
\begin{pmatrix}
0\\ 1 \\ 1 \\ 0 
\end{pmatrix},
\ \ \ \ \ \ 
-1:
\begin{pmatrix}
1\\ 0 \\ 0 \\ -1 
\end{pmatrix},
\begin{pmatrix}
0\\ 1 \\ -1 \\ 0 
\end{pmatrix}.
$$  

Problem 2 refers to an earlier problem that calculates the eigenvalues and eigenvectors of the matrix 
$$
\sigma_1=
\begin{pmatrix}
0&1 \\
1&0
\end{pmatrix}.

$$
[Edit by SN:] Added the full problem text. 

A:

The end result will be, of course, the same, but my view point is a bit different from those expressed by Arturo and Srivatsan, so here comes.
That matrix is giving us a linear mapping $T$ from a 4-dimensional vector space $U$ to itself.
Furthermore, we can express $U$ as a direct sum of its subspaces $U=V\oplus W$ in such a way that $T(V)\subseteq V$ and $T(W)\subseteq W$. Here $V$ is spanned by the first and fourth (standard) basis vectors of $U$, and $W$ is similarly spanned by the second and third basis vectors. In this sense $T$ is certainly a direct sum of its restrictions to these two complementary subspaces!
It is, perhaps a bit unusual that we don't order the basis vectors in such a way that the basis vectors belonging to one summand would come before those of the other. But, remember that the ordering of basis vectors is more or less arbitrary. Their indices are often just placeholders and/or a notational necessity.

