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Worked solution: Huang's signed hypercube adjacency matrix proof of the sensitivity conjecture (2019)

Step 5 of 7: Combining the pieces: the main theorem falls out
In plain words

Now the three previous ingredients snap together. The matrix AnA_n has n\sqrt{n} as an eigenvalue with huge multiplicity 2n−12^{n-1}; interlacing says any large enough principal submatrix must inherit an eigenvalue at least that big; and the row-sum fact converts that eigenvalue back into a statement about the maximum degree of the actual graph HH.

λ1(AH)≥λ2n−1(An)=n\lambda_1(A_H) \ge \lambda_{2^{n-1}}(A_n) = \sqrt{n}
Detailed analysis

Fix any induced subgraph HH of QnQ_n on a vertex set SS with ∣S∣≥2n−1+1|S| \ge 2^{n-1}+1, and let AHA_H be the principal submatrix of AnA_n obtained by keeping only the rows and columns indexed by SS. By Lemma 2.2 (Step 3), AnA_n has n\sqrt{n} as an eigenvalue of multiplicity 2n−12^{n-1}, which exceeds 2n−∣S∣2^n - |S|; applying Cauchy's interlacing theorem (Step 2) with i=2n−1i = 2^{n-1} therefore gives λ1(AH)≥λ2n−1(An)=n\lambda_1(A_H) \ge \lambda_{2^{n-1}}(A_n) = \sqrt{n}. Combining this with the row-sum lemma (Step 4) applied to AHA_H yields Δ(H)≥n\Delta(H) \ge \sqrt{n} -- exactly Huang's Theorem 1.1.

Knowledge used in this step