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The finite centering matrix is the orthogonal projection onto the zero-sum subspace

Statement

For $n \geq 1$ , let $H_{n} = I_{n} - n^{- 1} 1_{n} 1_{n}^{T}$ . Then $H_{n}^{T} = H_{n}$ , $H_{n}^{2} = H_{n}$ , $H_{n} 1_{n} = 0$ , and for each $v \in R^{n}$ , $H_{n} v = v - (n^{- 1} 1_{n}^{T} v) 1_{n}$ . Thus $H_{n}$ is the orthogonal projection onto ${v \in R^{n} : 1_{n}^{T} v = 0}$ .

Statement references

Hastie, Tibshirani, and Friedman, ESL II, §14.5.1, printed pp. 534–541

Cameron, Linear and Nonlinear Dimensionality Reduction, §2.2, PDF p. 4

Facts & Assumptions

Given: An integer $n \geq 1$ and $H_{n} = I_{n} - n^{- 1} 1_{n} 1_{n}^{T}$ .

[F1]

The finite centering matrix is $H_{n} = I_{n} - n^{- 1} 1_{n} 1_{n}^{T}$ , and it acts by subtracting the mean coordinate times $1_{n}$ (Finite centering matrices and centered data).

[F2]

The standard inner product on $R^{n}$ is $⟨ u, v ⟩ = \sum_{i < n} u_{i} v_{i}$ (The standard formulas $⟨ x, y ⟩ = \sum_{k < n} x_{k} y_{k}$ on $R^{n}$ and $\sum_{k < n} x_{k} \overline{y_{k}}$ on $C^{n}$ are inner products).

Proof

1.1givenF1algebra

The matrix $1_{n} 1_{n}^{T}$ is symmetric, so the formula in [F1] gives $H_{n}^{T} = H_{n}$ . Also $1_{n}^{T} 1_{n} = n$ , and hence $H_{n} 1_{n} = 1_{n} - n^{- 1} 1_{n} n = 0$ .

2.1step 1.1F1F2algebra∎

Writing $J = 1_{n} 1_{n}^{T}$ , one has $J^{2} = 1_{n} (1_{n}^{T} 1_{n}) 1_{n}^{T} = n J$ , so $H_{n}^{2} = (I_{n} - J / n)^{2} = I_{n} - 2 J / n + J^{2} / n^{2} = I_{n} - J / n = H_{n}$ . The formula for $H_{n} v$ in [F1] shows that its image is zero-sum; if $1_{n}^{T} v = 0$ , then $H_{n} v = v$ . For any $v$ , the residual $v - H_{n} v = (n^{- 1} 1_{n}^{T} v) 1_{n}$ is orthogonal to every zero-sum vector by [F2]. Hence $H_{n} v$ is the orthogonal projection of $v$ onto the zero-sum subspace.

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Sources

The Elements of Statistical Learning, second edition, corrected 12th printing (standard reference, not scraped)
Maria Cameron, Linear and Nonlinear Dimensionality Reduction, lecture notes (standard reference, not scraped)