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  "body": "\n## Proof\n\nTo show the tighter bound above, we instead consider the vectors $1_S-\\frac{|S|}n\\mathbf 1$ and $1_T-\\frac{|T|}n\\mathbf 1$, which are both perpendicular to $v_1$. We can expand\n\n$$1_S^\\operatorname{T}A_G1_T=\\left(\\frac{|S|}n\\mathbf 1\\right)^\\operatorname{T}A_G\\left(\\frac{|T|}n\\mathbf 1\\right)+\\left(1_S-\\frac{|S|}n\\mathbf 1\\right)^\\operatorname{T}A_G\\left(1_T-\\frac{|T|}n\\mathbf 1\\right)$$\n\nbecause the other two terms of the expansion are zero. The first term is equal to $\\frac{|S||T|}{n^2}\\mathbf 1^\\operatorname{T}A_G\\mathbf 1=\\frac dn|S||T|$, so we find that\n\n$$\\left|e(S,T)-\\frac dn|S||T|\\right| \\leq\\left|\\left(1_S-\\frac{|S|}n\\mathbf 1\\right)^\\operatorname{T}A_G\\left(1_T-\\frac{|T|}n\\mathbf 1\\right)\\right|$$\n\nWe can bound the right hand side by $\\lambda\\left\\|1_S-\\frac{|S|}{|n|}\\mathbf 1\\right\\|\\left\\|1_T-\\frac{|T|}{|n|}\\mathbf 1\\right\\| =\\lambda\\sqrt{|S||T|\\left(1-\\frac{|S|}n\\right)\\left(1-\\frac{|T|}n\\right)}$ using the same methods as in the earlier proof.\n",
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        "locator": "Expander mixing lemma, section Proofs \u003e Proof Sketch of Tighter Bound",
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