Proof two properties of the adjacency matrix of a graph
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Proof that for a graph G of order n the number of edges is $|E(G)|= frac{1}{2}trace(A^2)$ where A is the adjacency matrix of G.
Proof that for a simple graph G the number of triangles is given by the formula $frac{1}{6}trace(A^3)$
Any help?
graph-theory
asked Dec 22 '18 at 17:23
PCNFPCNF
1338
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add a comment |
$begingroup$
Proof that for a graph G of order n the number of edges is $|E(G)|= frac{1}{2}trace(A^2)$ where A is the adjacency matrix of G.
Proof that for a simple graph G the number of triangles is given by the formula $frac{1}{6}trace(A^3)$
Any help?
graph-theory
asked Dec 22 '18 at 17:23
PCNFPCNF
1338
$endgroup$
add a comment |
$begingroup$
Proof that for a graph G of order n the number of edges is $|E(G)|= frac{1}{2}trace(A^2)$ where A is the adjacency matrix of G.
Proof that for a simple graph G the number of triangles is given by the formula $frac{1}{6}trace(A^3)$
Any help?
graph-theory
asked Dec 22 '18 at 17:23
PCNFPCNF
1338
$endgroup$
Proof that for a graph G of order n the number of edges is $|E(G)|= frac{1}{2}trace(A^2)$ where A is the adjacency matrix of G.
Proof that for a simple graph G the number of triangles is given by the formula $frac{1}{6}trace(A^3)$
Any help?
graph-theory
graph-theory
asked Dec 22 '18 at 17:23
PCNFPCNF
1338
asked Dec 22 '18 at 17:23
PCNFPCNF
1338
asked Dec 22 '18 at 17:23
PCNFPCNF
1338
asked Dec 22 '18 at 17:23
PCNFPCNF
1338
asked Dec 22 '18 at 17:23
PCNFPCNF
1338
1338
add a comment |
add a comment |
1 Answer
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Note that the $ij$ entry of $A^n$ counts the number of $i-j$ walks of length $n$ in the graph. The $i-i$ walks are closed. When $n=2$, the diagonal entries count vertex degrees. Each edge is counted twice. Hence, the first result.
When $n=3$, the closed walks are triangles. There are three vertices in a triangle. The triangle is counted twice by each vertex (circle forwards, circle backwards). Hence, we divide by 6. This yields your second result.
answered Dec 22 '18 at 17:28
ml0105ml0105
11.5k21538
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$begingroup$
Note that the $ij$ entry of $A^n$ counts the number of $i-j$ walks of length $n$ in the graph. The $i-i$ walks are closed. When $n=2$, the diagonal entries count vertex degrees. Each edge is counted twice. Hence, the first result.
When $n=3$, the closed walks are triangles. There are three vertices in a triangle. The triangle is counted twice by each vertex (circle forwards, circle backwards). Hence, we divide by 6. This yields your second result.
answered Dec 22 '18 at 17:28
ml0105ml0105
11.5k21538
$endgroup$
add a comment |
$begingroup$
Note that the $ij$ entry of $A^n$ counts the number of $i-j$ walks of length $n$ in the graph. The $i-i$ walks are closed. When $n=2$, the diagonal entries count vertex degrees. Each edge is counted twice. Hence, the first result.
When $n=3$, the closed walks are triangles. There are three vertices in a triangle. The triangle is counted twice by each vertex (circle forwards, circle backwards). Hence, we divide by 6. This yields your second result.
answered Dec 22 '18 at 17:28
ml0105ml0105
11.5k21538
$endgroup$
add a comment |
$begingroup$
Note that the $ij$ entry of $A^n$ counts the number of $i-j$ walks of length $n$ in the graph. The $i-i$ walks are closed. When $n=2$, the diagonal entries count vertex degrees. Each edge is counted twice. Hence, the first result.
When $n=3$, the closed walks are triangles. There are three vertices in a triangle. The triangle is counted twice by each vertex (circle forwards, circle backwards). Hence, we divide by 6. This yields your second result.
answered Dec 22 '18 at 17:28
ml0105ml0105
11.5k21538
$endgroup$
Note that the $ij$ entry of $A^n$ counts the number of $i-j$ walks of length $n$ in the graph. The $i-i$ walks are closed. When $n=2$, the diagonal entries count vertex degrees. Each edge is counted twice. Hence, the first result.
When $n=3$, the closed walks are triangles. There are three vertices in a triangle. The triangle is counted twice by each vertex (circle forwards, circle backwards). Hence, we divide by 6. This yields your second result.
answered Dec 22 '18 at 17:28
ml0105ml0105
11.5k21538
answered Dec 22 '18 at 17:28
ml0105ml0105
11.5k21538
answered Dec 22 '18 at 17:28
ml0105ml0105
11.5k21538
answered Dec 22 '18 at 17:28
ml0105ml0105
11.5k21538
11.5k21538
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