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So I am having a problem with a following problem:

Let there be matrix $J\in\mathbb{R}^{n\times n}$, which has number $1$ everywhere. Solve the equation for $X$: $J-X=JX$.

I think I have to apply a trace of a matrix, but I am not sure how, since you have a product of matrices on the right side.

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Hint: Rewrite the equation as $$ (J+I)X=J $$ and show that $\det(J+I)\neq 0$, so that $$ X=(J+I)^{-1}J. $$ There is no need of trace.

The determinant of a matrix with all non-diagonal coefficients equal to $1$ and diagonal $2$ has been computed at this site, see for example here:

Determinant of a specially structured matrix ($a$'s on the diagonal, all other entries equal to $b$)

Finding determinant for a matrix with one value on the diagonal and another everywhere else

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You may also write $J=ee^T$, where $e^T=(1,1,\ldots,1)$. The equation is then equivalent to $ee^T-X=ee^TX$ or $X=e(e^T-e^TX)$. Let $v^T=e^T-e^TX$. Then $X=ev^T$ and $v^T=e^T-e^TX=e^T-e^Tev^T=e^T-nv^T$. Hence $v^T=\frac{1}{n+1}e^T$ and $X=ev^T=\frac{1}{n+1}ee^T=\frac{1}{n+1}J$.

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  • $\begingroup$ In what form is $v^T$ here? $\endgroup$ – david1201 Feb 23 at 11:05
  • $\begingroup$ @david1201 What do you mean? $\endgroup$ – user1551 Feb 23 at 11:10
  • $\begingroup$ I am not sure why did you use $v^T$ and how does this matrix looks? Sorry, I have just started with learning linear algebra this week. $\endgroup$ – david1201 Feb 23 at 11:15
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    $\begingroup$ @david1201 $v^T$ is defined to be $e^T-e^TX$. So, it's a row vector. $\endgroup$ – user1551 Feb 23 at 11:17

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