1. Spectral Sparsifiers
1.1. Graph Laplacians
Let be an unweighted graph. For notational simplicity, we will think of the vertex set as . Let be the th standard basis vector, meaning that has a in the th coordinate and s in all other coordinates. For an edge , define the vector and the matrix as follows:
In the definition of it does not matter which vertex gets the and which gets the because the matrix is the same either way.
Definition 1 The Laplacian matrix of is the matrix
Let us consider an example.
Note that each matrix has only four non-zero entries: we have and . Consequently, the th diagonal entry of is simply the degree of vertex . Moreover, we have the following fact.
Fact 2 Let be the diagonal matrix with equal to the degree of vertex . Let be the adjacency matrix of . Then .
If had weights on the edges we could define the weighted Laplacian as follows:
Claim 3 Let be a graph with non-negative weights . Then the weighted Laplacian is positive semi-definite.
Proof: Since , it is positive semi-definite. So is a weighted sum of positive semi-definite matrices with non-negative coefficients. Fact 5 in the Notes on Symmetric Matrices implies is positive semi-definite.
The Laplacian can tell us many interesting things about the graph. For example:
Proof: For any edge we have . But is if exactly one of or is in , and otherwise it is . So if , and otherwise it is . Summing over all edges proves the claim.
Similarly, if is a graph with edge weights and is the weighted Laplacian, then then .
1.2. Main Theorem
- only of the weights are non-zero, and
- with probability at least ,
where denotes the weighted Laplacian of with weights . By Fact 4 in Notes on Symmetric Matrices, this is equivalent to
The algorithm that proves Theorem 6 is as follows.
- Initially .
- Set .
- For every edge compute .
- Let be a random edge chosen with probability .
- Increase by .
Proof: (of Theorem 6). How does the matrix change during the th iteration? The edge is chosen with probability and then increases by . Let be this random change in during the th iteration. So equals with probability . The random matrices are mutually independent and they all have this same distribution. Note that
The final matrix is simply . To analyze this final matrix, we will use the Ahlswede-Winter inequality. All that we require is the following claim, which we prove later.
We apply Corollary 2 from the previous lecture with , obtaining
2. Appendix: Additional Proofs
Proof: (of Claim 7) First we check that the values are non-negative. By the cyclic property of trace
This is non-negative since because . Thus . Next, note that
where is the orthogonal projection onto the image of . The image has dimension by Fact 5, and so
Proof: (of Claim 8). The maximum eigenvalue of a positive semi-definite matrix never exceeds its trace, so
By Fact 8 in the Notes on Symmetric Matrices,
So, by Fact 4 in the Notes on Symmetric Matrices, for every vector ,
Now let us write where is the projection onto the image of and is the projection onto the kernel of . Then and . So
Since this holds for every vector , Fact 4 in the Notes on Symmetric Matrices again implies
Since , Claim 16 in the Notes on Symmetric Matrices shows this is equivalent to
This completes the proof of the claim.