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Suppose I have a matrix where $x \in \mathbb{Z}$

$ A = \begin{bmatrix} x_{11} & x_{12} & x_{13} & \dots & x_{1n} \\ x_{21} & x_{22} & x_{23} & \dots & x_{2n} \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ x_{d1} & x_{d2} & x_{d3} & \dots & x_{dn} \end{bmatrix}$

what is the best way to arrive at the following matrix:

$ B = \begin{bmatrix} \frac{1}{x_{11}} & \frac{1}{x_{12}} & \frac{1}{x_{13}} & \dots & \frac{1}{x_{1n}} \\ \frac{1}{x_{21}} & \frac{1}{x_{22}} & \frac{1}{x_{23}} & \dots & \frac{1}{x_{2n}} \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ \frac{1}{x_{d1}} & \frac{1}{x_{d2}} & \frac{1}{x_{d3}} & \dots & \frac{1}{x_{dn}} \end{bmatrix}$

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    $\begingroup$ Find the reciprocal of each element. $\endgroup$ – Kenny Lau Aug 2 '16 at 8:22
  • $\begingroup$ yes...how would you write this? $\endgroup$ – TsTeaTime Aug 2 '16 at 8:22
  • $\begingroup$ "Find the reciprocal of each element." $\endgroup$ – Kenny Lau Aug 2 '16 at 8:22
  • $\begingroup$ I don't suppose i can write $1/A$ $\endgroup$ – TsTeaTime Aug 2 '16 at 8:23
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    $\begingroup$ "$B$ is the matrix of the reciprocal of each element in $A$" $\endgroup$ – Kenny Lau Aug 2 '16 at 8:24
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You can define the Schur product (sometimes also called Hadamard product) as follows:

If $A, B \in \mathbb K^{n \times m}$ then the Schurproduct $A \ast B$ is defined as $(A \ast B)_{i,j} = (A)_{i,j} (B)_{i,j}$, i.e. the multiplication of the entries.

Using that you can say the following:

"Let $B$ the inverse of $A$ respecting the Schur product."

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  • $\begingroup$ Wonderful...thank you for this @Yaddle. This is really helpful! $\endgroup$ – TsTeaTime Aug 2 '16 at 8:46
  • $\begingroup$ Does $\mathbb K^{n \times m}$ represent anything in particular? i am not able to find this under the list of Number Sets? $\endgroup$ – TsTeaTime Aug 2 '16 at 8:49
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    $\begingroup$ In my notation you have $\mathbb K \in \{\mathbb R, \mathbb C\}$ and $\mathbb K^{n \times m}$ denotes the space of matrices with $n$ rows and $m$ columns with entries in $\mathbb K$. Hope that makes it a little clearer :) $\endgroup$ – Yaddle Aug 2 '16 at 9:19

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