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I'm dealing with the following problem:

Suppose that I gave a upper diagonal matrix A of the form: $A= \begin{bmatrix} a_{11} & a_{12}&\dots &a_{1n}\\ 0 & a_{22}&\dots &a_{2n} \\ \vdots & \vdots&\dots &\vdots\\ 0 & 0&\dots &a_{nn} \end{bmatrix}$

And a matrix $A'$ that is $A$ with only the diagonal elements multiplied by values $u_1,u_2,\dots,u_n$, but the upper diagonal elements stay intact, that is: $A' = \begin{bmatrix} u_1 a_{11} & a_{12}&\dots &a_{1n}\\ 0 & u_2 a_{22}&\dots &a_{2n} \\ \vdots & \vdots&\dots &\vdots\\ 0 & 0&\dots &u_n a_{nn} \end{bmatrix}$

How can I express $A'$ in terms of $A$?

Is there a way to express it in the form $A' =UA$? But then, what is the form of the matrix U?

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  • $\begingroup$ $A' = \begin{bmatrix} u_{1} & 0&\dots &0\\ 0 & u_{2}&\dots &0 \\ \vdots & \vdots&\dots &\vdots\\ 0 & 0&\dots &u_{n} \end{bmatrix}\begin{bmatrix} a_{11} & 0&\dots &0\\ 0 & a_{22}&\dots &0 \\ \vdots & \vdots&\dots &\vdots\\ 0 & 0&\dots &a_{nn} \end{bmatrix}+\begin{bmatrix} 0 & a_{12}&\dots &a_{1n}\\ 0 & 0&\dots &a_{2n} \\ \vdots & \vdots&\dots &\vdots\\ 0 & 0&\dots &0 \end{bmatrix}$. You got to isolate the diagonal elements and then multiply I guess. $\endgroup$ – Yadati Kiran Nov 22 '18 at 17:45
  • $\begingroup$ Just calculate $U=A'A^{-1}$ $\endgroup$ – Widawensen Nov 22 '18 at 17:51
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    $\begingroup$ How about $$\eqalign{ A' &= A + {\rm Diag}(u\!-\!1){\,\rm Diag}(A) \cr U &= A'A^{-1} = I + {\rm Diag}(u\!-\!1){\,\rm Diag}(A)A^{-1} \cr }$$ $\endgroup$ – greg Nov 22 '18 at 18:01
  • $\begingroup$ @Widawensen I need something that does not involve $A'$ $\endgroup$ – A.T Nov 22 '18 at 18:08
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    $\begingroup$ @ArthurT Anyway, the result is diag$(u_1,u_2,u_3)$ + some nilpotent matrix with more complicated entries. $\endgroup$ – Widawensen Nov 22 '18 at 18:19

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