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<\/figure>\n\n\n\nTo invert the matrix elementwise, we represent it as in the figure above, as consisting of the sum of a diagonal matrix and a ‘corner’ matrix made of the same vector $h$ as the leftmost column and the topmost row. That is,
$$ M_{ij} = \\delta_{ij} + \\delta_{i1}h_j + h_i\\delta_{1j} – h_{1} \\delta_{i1}\\delta_{1j},$$ where the last term avoids double counting the term at the top-left corner.<\/p>\n\n\n\n
The inverse $G$ satisfies $\\sum_{k} M_{ik} G_{kj} = \\delta_{ij}.$ Then \\begin{align}\\delta_{ij} &= \\sum_k M_{ik} G_{kj}\\\\
&= \\sum_k (\\delta_{ik} + h_i \\delta_{1k} + \\delta_{i1}h_k – \\delta_{i1}\\delta_{1k}h_{1})G_{kj}\\\\
&= G_{ij} + h_i G_{1j} + \\delta_{i1}\\sum_k (h_k – \\delta_{1k} h_{1})G_{kj} \\\\
&= G_{ij} + h_i G_{1j} + \\delta_{i1}\\left(-h_{1} G_{1j} + \\sum_k h_k G_{kj}\\right)\\\\&= G_{ij} + h_i G_{1j} + \\delta_{i1} \\sum_{k>1} h_k G_{kj}.
\\end{align}<\/p>\n\n\n\n
Let’s first consider the case where $i>1$. Then
\\begin{align} G_{ij} &= \\delta_{ij} – h_i G_{1j}.
\\end{align} Let’s define $g_j = G_{1j}$. Then rows below the top look like $I – h g^T$.<\/p>\n\n\n\n
Now let’s consider the case where $i=1$. Then
\\begin{align}
\\delta_{1j} &= g_j + h_1 g_j + \\sum_{k>1} h_k G_{kj}\\\\
&= (1 + h_1)g_j + \\sum_{k>1}h_k (\\delta_{kj} – h_k g_j)\\\\&= (1 + h_1 + h_1^2 – \\|h\\|_2^2)g_j + (j>1) h_j.\\end{align} So, defining $$h_0 \\triangleq 1 + h_1 + h_1^2 – \\|h\\|_2^2$$ we get \\begin{align} g_1 &= {1 \\over h_0}, \\quad g_{k>1} = -{h_k \\over h_0}.\\end{align} Putting it altogether we arrive at
$$G = I – e_1 e_1^T + ((1+h_1)e_1 – h)g^T.$$ The $I – e_1 e_1^T$ term covers the required identity term when $i>1$, Here the $e_1 e_1^T$ terms subtracts the top-left 1 from the identity matrix, and the coefficients of $g^T$ are such that the first is $1$, and the remainder are $-h_2, -h_3, \\dots$ <\/p>\n\n\n\n
In fact, we notice that the latter vector forms the numerator of $g$. Therefore, we assign it its own symbol
$$\\tilde h \\triangleq [1, -h_2, -h_3, ..],$$ in terms of which $g = \\tilde h\/h_0,$ and
$$ \\boxed{G = I – e_1 e_1^T + {\\tilde h \\tilde h^T \\over h_0}.}$$<\/p>\n\n\n\n