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I am trying to prove the following inequality:

$$ x^T(I_d-11^T/d)x \ge 0 $$

where $d$ is a non-negative integer, $I_d$ a $d\times d$ identity matrix and $1$ a $d \times 1$ vector composed of ones. I can prove this by using the lemma that for any real symmetric $d \times d$ matrix $A$ the following holds:

$$ \lambda_{(1)} \le \frac{x^TAx}{x^Tx} \le \lambda_{(d)} $$

where $\lambda_{(1)}$ is the smallest and $\lambda_{(d)}$ the largest eigenvalue. Combining this in the above equation one can show (if I am not mistaken) that

$$ 0 \le x^T(I_d-11^T/d)x \le x^Tx $$

which proves the semi-positive definiteness.

Are there any more intuitive ways to prove this without the appeal to this lemma?

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    The matrix $I_d-11^T/d$ is a projection matrix (to the plane $\sum_{k=1}^d x_k=0$ specifically). So this is an instance of the fact that projection matrices are positive semi-definite. To see why this should be true, consider the angle between a vector and its vector projection onto a given plane. – 2017-02-06
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    @Semiclassical very good comment that could have been an answer... – 2017-02-06

2 Answers 2

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Sure. Let $v$ denote the unit-vector $1/\sqrt{d}$. Then $$ (I - 11^T/d) = (I - vv^T) $$ It follows that $$ x^T(I - 11^T/d)x = x^T(I - vv^T)x^T = x^Tx - (x^Tv)^2 =\\ \|x\|^2 - |\langle x,v \rangle|^2 $$ and by Cauchy-Schwarz, this must be non-negative.

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Note that

$$\mathrm P := \frac{1}{d} 1_d 1_d^\top = \left( \frac{1}{\sqrt{d}} 1_d \right) \left( \frac{1}{\sqrt{d}} 1_d \right)^\top$$

is a symmetric, rank-$1$ (projection) matrix whose eigenvalues are $1$ and $0$, with multiplicities $1$ and $d-1$, respectively. Thus, $\mathrm I_d - \mathrm P$ is a symmetric (projection) matrix whose eigenvalues are $0$ and $1$, with multiplicities $1$ and $d-1$, respectively. Hence, matrix $\mathrm I_d - \mathrm P$ is positive semidefinite and

$$\mathrm x^\top \left( \mathrm I_d - \mathrm P \right) \,\mathrm x \geq 0$$

holds for all $\mathrm x \in \mathbb R^d$.