Let $A_{n \times n}$ be a positive-definite matrix. There is a way to approximate the following product?
$$\left(I-\frac{1}{n}11^{T}\right) A \left(I-\frac{1}{n}11^{T}\right)$$
where $1$ is a column vector of ones.
Let $A_{n \times n}$ be a positive-definite matrix. There is a way to approximate the following product?
$$\left(I-\frac{1}{n}11^{T}\right) A \left(I-\frac{1}{n}11^{T}\right)$$
where $1$ is a column vector of ones.
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Let $U_1:=\dfrac{1}{\sqrt{n}}\boldsymbol{1}$ the normalized version of $\boldsymbol{1}$.
Let $\Pi=\mathbb{R}U_1^{\perp}$ (hyperplane orthogonal to vector $V$).
Let $\{U_2,U_3,\cdots U_n\}$ be vectors complementing $U_1$ in such a way that they constitute an ortho-normal basis of $\mathbb{R^n}$ (each vector is orthogonal to all others ; all with unit norms).
Now comes the geometric interpretation :
$U_1U_1^T$ is tha matrix of the orthogonal projection onto one-dimensional subspace $\mathbb{R}U_1$.
Therefore, $$P:=I-\frac1n \boldsymbol{11}^T=I-U_1U_1^T$$ is the matrix of the orthogonal projection onto hyperplane $\Pi$.
As a consequence, $P^T=P$ ;
The computation of the value of the quadratic form
$$X^T(PAP)X=(PX)^TA(PX)$$
is based on the efficiency of computation of $PX$.
This can be done by expressing the classical decomposition of a vector onto a subspace and its complementary subspace :
$$X=U_1(U_1^T.X)+V$$ where $V \in \Pi$, giving
$$PX=\underbrace{PU_1}_{= \ 0}(U_1^T.X)+\underbrace{PV}_{= \ V}.$$