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Let X is nxp (design) matrix and $H=X({X^\top}X)^{-1}{X^\top}$. We define $h_{ij}$ is an i-th row and j-th column entry of H and $x_i^\top$ is i-th row of $X$ , then show $$h_{ii}={x_i^\top}({X^\top}X)^{-1}x_i$$ and $$h_{ji}={x_j^\top}({X^\top}X)^{-1}x_i$$

I tried with definition of matrix multiplication of each entry, but was not straight forward. Anyone who has better idea or can solve with my approach?

Thanks in advance!

Horselook
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1 Answers1

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For any matrix $A$, its $ij$-th entry is given by

$ a_{ij} = e_i^T A e_j $

where $e_k$ is $k$-th column of the identity matrix. Hence,

$ h_{ij} = e_i^T X (X^T X)^{-1} X^T e_j $

Recognizing that $e_i^T X = ( X^T e_i )^T $ and $X^T e_i $ is the $i$-th column of $X^T$ which is the $i$-th row of $X$, which is $x_i$, hence

$h_{ij} = x_i^T(X^T X)^{-1} x_j $

And letting $j = i$ then

$h_{ii} = x_i^T (X^T X)^{-1} x_i $

Hosam Hajeer
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