Is it true that $ x_i (X^TX)^{-1} x_i^T = \frac{1}{n} + \frac{(x_i - \bar{x})(x_i - \bar{x})^T}{\sum_{j=1}^n(x_j - \bar{x})(x_j - \bar{x})^T}$?

Question

Suppose $X$ is a $n\times m$ matrix. And also define $x_i$ to be (part of) the $i$-th column of $X$:

$$ X = \begin{bmatrix} 1 & 1 & ... & 1 \\ x_1 & x_2 & ... & x_n \\ \end{bmatrix} $$

Is it true that: $$ x_i (X^TX)^{-1} x_i^T = \frac{1}{n} + \frac{(x_i - \bar{x})(x_i - \bar{x})^T}{\sum_{j=1}^n(x_j - \bar{x})(x_j - \bar{x})^T} $$

In a sense it all makes sense, since $X^TX$ should act like normalization to the whole thing and trying the special cases it works. Although I don't find a way to prove it for general case. One big difficulty is that I don't know how to model the matrix inverse $(X^TX)^{-1}$. Any ideas?

Update: another form of this is implied in this work (equation 3.1), where it defines: $$H = X (X^TX)^{-1} X^T$$

where $h_{ii}$ is the $i$-th diagonal element of this matrix.

Answer 1

Not true. Let $X=I$ be a $n\times n$ identity matrix. Then $X^TX = I$. The left hand side is $x_ix_i^T = 1$. While the RHS is $\frac{1}{n}$ since $(x_i -\bar{x})(x_i- \bar{x})^T$ is equal for all $i=1,\ldots,n$.

You might also try $X=x$ a $1\times m$ matrix. Then $\bar{x}=x$. So the RHS is zero, while the LHS is not, since it has nonzero trace ($\text{Tr}(x(x^Tx)^{-1}x^T)=\text{Tr}((x^Tx)^{-1}x^Tx)=\text{Tr}(I_{m\times m})=m$.