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Let $A$ and $B$ be square matrices of same dimension. I considered $n$-by-$n$ block matrices of the form

\begin{align*} \begin{pmatrix} A & & \\ & \ddots & \\ & & A \end{pmatrix} + \begin{pmatrix} B & \dots & B \\ \vdots & \ddots & \vdots \\ B & \dots & B \end{pmatrix}. \end{align*}

I got somehow pointed into the direction that this block matrix is similar to a block triangular matrix with diagonal entries $A + nB$ one time and $A+B$ ($n-1$-times).

But I could not figure out an easy way to see this. Any help is greatly appreciated. Thanks.

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    $\begingroup$ Presumably you meant $A$ ($n-1$) times? $\endgroup$ – copper.hat May 28 '15 at 17:39
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Let $I$ denote the identity matrix and $x = (1,\dots,1)^T$. Let $\otimes$ denote the Kronecker product. This matrix can be written in the form $$ I \otimes A + (xx^T) \otimes B $$ We note that there is an orthogonal matrix $U$ such that $U(xx^T)U^T = nE_{11}$, where $$ E_{11} = \pmatrix{1&0&\cdots\\0&0\\\vdots&&\ddots} $$

It follows that

$$ (U \otimes I)(I \otimes A + (xx^T) \otimes B)(U \otimes I)^T =\\ I \otimes A + (U(xx^T)U^T) \otimes B =\\ I \otimes A + n E_{11} \otimes B =\\ \pmatrix{ A+nB\\ &A\\ &&\ddots\\ &&&A } $$

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    $\begingroup$ One possibility for $U$ of dimension $m\times m$ would be $$\begin{pmatrix}1/\sqrt m & 1/\sqrt 2 & 1/\sqrt 6 & 1/\sqrt{12} & \cdots & 1/\sqrt{m(m-1)} \\ 1/\sqrt m & -1/\sqrt 2 & 1/\sqrt 6 & 1/\sqrt{12} & \cdots & 1/\sqrt{m(m-1)} \\ 1/\sqrt m & 0 & -2/\sqrt 6 & 1/\sqrt{12} & \cdots & 1/\sqrt{m(m-1)} \\ 1/\sqrt m & 0 & 0 & -3/\sqrt{12} & & 1/\sqrt{m(m-1)} \\ \vdots & \vdots & \vdots & \vdots & \ddots & \vdots \\ 1/\sqrt m & 0 & 0 & 0 & \cdots & -(m-1)\sqrt{m(m-1)} \end{pmatrix} $$ $\endgroup$ – Henning Makholm May 28 '15 at 17:54
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    $\begingroup$ For some reason, I always find the Kronecker product a little intimidating. Perhaps a throw back from the days of 8kb computers :-). $\endgroup$ – copper.hat May 29 '15 at 4:13

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