I want to demonstrate that the centering matrix $H$ is idempotent (i.e. $HH=H$). The centering matrix is defined as $H=I-\frac{1}{n}1\, 1^T$.
I've tried developing this: $$ HH=\\ H(I-\frac{1}{n}1\, 1^T)= \\ HI-H\frac{1}{n}1\, 1^T=\\ I-\frac{1}{n}1\, 1^T-(I-\frac{1}{n}1\, 1^T)\frac{1}{n}1\, 1^T= \\ I-\frac{1}{n}1\, 1^T-\frac{1}{n}1\, 1^T+\frac{1}{n^2}1\, 1^T $$
From here, I should obtain again the original form $I-\frac{1}{n}1\, 1^T$, but I don't know how. Any ideas on how to proceed? Or any ideas on a different approach?
In the third line you have $HI - H \frac{1}{n} 11^T$. So the assertion follows if $H \frac{1}{n} 11^T = 0$. Now substitute again for $H$, you'll see that the assertion follows if $11^T = \frac{1}{n} 11^T 11^T$. In $\mathbb{R}^n$ we have $11^T 11^T = n 11^T$ so the assertion follows.
To see why $11^T 11^T = n 11^T$ note that $11^T_{i,j} = \sum_{i=1}^n 1 \times 1$.