adjacency matrixmatrices
I was unable to find a mathematical operation for obtaining the degree matrix from the adjacency matrix of a given graph.
For a graph $G = (V,E)$, let $A$ be the adjacency matrix of $G$ and let $D \in \mathbb{R}^{{|V|}\times |V|}$ be the (diagonal) degree matrix, $D = \mbox{diag} \left( A 1_{|V|} \right)$, where $1_{|V|}$ is the vector of all-ones of dimension $|V|$. More specifically, I want to know if we can mathematically define the $\mbox{diag}(\cdot)$ operation.
Theres a whole field known as "Spectral graph theory", where the eigendecomposition of of certain matrices (typically the Laplacian matrix, $L=D-A$ where $D$ is the diagonal matrix with the sum of degrees of each vertex) is studied. This is a lot more interesting than looking at the spectrum of the adjacency matrix for many reasons (for example, you can make connections to harmonic function theory).
One can prove that the largest eigenvalue of the Adjacency matrix lies between $max(d_{avg},\sqrt{d_{max}})$ and $d_{max}$ (see these notes).
I will give an example, and to make the typing as economical as possible, i will use sage:
sage: G = graphs.PetersenGraph()
sage: G
Petersen graph: Graph on 10 vertices
sage: G.adjacency_matrix()
[0 1 0 0 1 1 0 0 0 0]
[1 0 1 0 0 0 1 0 0 0]
[0 1 0 1 0 0 0 1 0 0]
[0 0 1 0 1 0 0 0 1 0]
[1 0 0 1 0 0 0 0 0 1]
[1 0 0 0 0 0 0 1 1 0]
[0 1 0 0 0 0 0 0 1 1]
[0 0 1 0 0 1 0 0 0 1]
[0 0 0 1 0 1 1 0 0 0]
[0 0 0 0 1 0 1 1 0 0]
sage: G.vertices()
[0, 1, 2, 3, 4, 5, 6, 7, 8, 9]
sage: G.show()
Launched png viewer for Graphics object consisting of 26 graphics primitives
Above, the rows and the columns of the matrix are labeled $0,1,2,3,4,5,6,7,8,9$ and considered with this order of them. For instance, $6$ and $9$ are joined by an edge, because of this we have the $\boxed 1$ entry in row $6$ column $9$ (and vice versa):
$$
\left[\begin{array}{rrrrrrrrrr}
0 & 1 & 0 & 0 & 1 & 1 & 0 & 0 & 0 & 0 \\
1 & 0 & 1 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \\
0 & 1 & 0 & 1 & 0 & 0 & 0 & 1 & 0 & 0 \\
0 & 0 & 1 & 0 & 1 & 0 & 0 & 0 & 1 & 0 \\
1 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 1 \\
1 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 1 & 0 \\
0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & \boxed 1 \\
0 & 0 & 1 & 0 & 0 & 1 & 0 & 0 & 0 & 1 \\
0 & 0 & 0 & 1 & 0 & 1 & 1 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 1 & 0 & \boxed 1 & 1 & 0 & 0
\end{array}\right]
\ .
$$
The first row is 0 1 0 0 1 1 0 0 0 0 corresponding to the columns $0,1,2,3,4,5,6,7,8,9$, and the ones are corresponding to $1,4,5$, which are the neighbours of $0$ in the graph.
Now it is possible that an alien prefers an other order, for instance
$9,6,7,2,1,3,5,4,0,8$. Then he/she/it should write an other matrix. The corresponding boxed entries are now
$$
\left[\begin{array}{rrrrrrrrrr}
0 & \boxed 1 & 1 & 0 & 0 & 0 & 0 & 1 & 0 & 0 \\
\boxed 1 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 1 \\
1 & 0 & 0 & 1 & 0 & 0 & 1 & 0 & 0 & 0 \\
0 & 0 & 1 & 0 & 1 & 1 & 0 & 0 & 0 & 0 \\
0 & 1 & 0 & 1 & 0 & 0 & 0 & 0 & 1 & 0 \\
0 & 0 & 0 & 1 & 0 & 0 & 0 & 1 & 0 & 1 \\
0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 1 & 1 \\
1 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 1 & 0 \\
0 & 0 & 0 & 0 & 1 & 0 & 1 & 1 & 0 & 0 \\
0 & 1 & 0 & 0 & 0 & 1 & 1 & 0 & 0 & 0
\end{array}\right]
$$
The first row is 0 1 1 0 0 0 0 1 0 0 corresponding to the columns $9,6,7,2,1,3,5,4,0,8$, and the ones are corresponding to $6,7,4$, which are the neighbours of $9$ in the graph.
The following code was used to get he new matrix for the new order.
sage: L = [9,6,7,2,1,3,5,4,0,8]
sage: A = G.adjacency_matrix()
sage: B = matrix(ZZ, 10, 10, [A[m,n] for m in L for n in L] )
sage: latex(B)
Note: The cited text is very pedant, puts the accent on some bureaucratic point, the order, then introduces a lot of notation for this, a permutation $\pi$, the space of permutations, makes it part of the structure. This is maybe misleading at least in the same measure i am in similar situations.
Best Answer
As with any linear mapping, the function $F:\mathbb R^n \to \mathbb R^{n\times n}$ that sends a vector $\vec v = (v_1,\ldots,v_n)$ to a diagonal matrix with those entries:
$$ F\, \vec v = \begin{pmatrix} v_1 & & 0 \\ & \ddots & \\ 0 & & v_n \end{pmatrix} $$
is uniquely determined by its action on a basis. With a suitable choice of notation we can turn such a definition into a "formula".
Recall the standard ordered basis for $\mathbb R^n$, namely row vectors:
$$ \textbf e_1 = (1,0,\ldots,0), \ldots, \textbf e_n = (0,\ldots,0,1) $$
We can then define our linear mapping as a sum:
$$ F(v_1,\ldots,v_n) = \sum_{k=1}^n v_k \textbf e_k^T \textbf e_k $$
Note that the product of column vector $\textbf e_k^T$ and the row vector $\textbf e_k$ gives a diagonal $n\times n$ matrix with $1$ in the $k$th diagonal position (and zeros elsewhere). Therefore the summation of these terms gives a diagonal matrix supplying exactly the specified diagonal entries.